9.MECHANICAL PROPERTIES OF SOLIDSQ. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10–5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load.What is the ratio of the Young’s modulus of steel to that of copper?AnswerLength of the steel wire, L1 = 4.7 mArea of cross-section of the steel wire, A1 = 3.0 × 10–5 m2Length of the copper wire, L2 = 3.5 m A2 = 4.0 × 10–5 m2 Change in length = L1 = L2 = LArea of cross-section of the copper wire,Force applied in both the cases = F Young’s modulus of the steel wire:Y1 = (F1 / A1) (L1 / L1)= (F / 3 X 10-5) (4.7 / L) ....(i) Young’s modulus of the copper wire: Y2 = (F2 / A2) (L2 / L 2)= (F / 4 × 10-5) (3.5 / L) ....(ii) Dividing (i) by (ii), we get:Y1 / Y2 = (4.7 × 4 × 10-5) / (3 × 10-5 × 3.5)= 1.79 : 1The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.Q.. Figure 9.11 shows the strain-stress curve for a given material. What are(a) Young’s modulus and(b) approximate yield strength for this material?Ans(a) It is clear from the given graph that for stress 150 × 106N/m2, strain is 0.002.∴Young’s modulus, Y = Stress / Strain = 150 × 106 / 0.002 = 7.5 × 1010 Nm-2Hence, Young’s modulus for the given material is 7.5 ×1010 N/m2.(b) The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit.It is clear from the given graph that the approximate yield strength of this materialIs 300 × 106 Nm/2 or 3 × 108 N/m2.Q.. The stress-strain graphs for materials A and B are shown in Fig. 9.12.The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus? (b) Which of the two is the stronger material? Ans. (a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence, (b) Young's modulus (b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture. 9.4. Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young’s modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus. Ans. (a) False, because for given stress there is more strain in rubber than steel and modulus of elasticity is (b) inversely proportional to strain. (c) True, because the stretching of coil simply changes its shape without any change in the length of the (d) wire used in the coil due to which shear modulus of elasticity is involved. Q. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.AnswerElongation of the steel wire = 1.49 × 1100––44 m Diameter of the wires, d = 0.25 mElongation of the brass wire = 1.3 × mHence, the radius of the wires, r = d/2 = 0.125 cm Length of the steel wire, L1 = 1.5 mLength of the brass wire, L2 = 1.0 mTotal force exerted on the steel wire: F1 = (4 + 6) g = 10 × 9.8 = 98 N Young’s modulus for steel:Y1 = (F1/A1) / ( L1 / L1) Where,L1 = Change in the length of the steel wireA1 = Area of cross-section of the steel wire = πr12Young’s modulus of steel, Y1 = 2.0 × 1011 Pa ∴ L1 = F1 × L1 / (A1 × Y1)= (98 × 1.5) / [ π(0.125 × 10-2)2 × 2 × 1011] = 1.49 × 10-4 mTotal force on the brass wire: F2 = 6 × 9.8 = 58.8 N
Young’s modulus for brass: Y2 = (F2/A2) / ( L2 / L2) Where, L2 = Change in the length of the brass wire A1 = Area of cross-section of the brass wire = πr12 ∴ L2 = F2 × L2 / (A2 × Y2) = (58.8 X 1) / [ (π × (0.125 × 10-2)2 × (0.91 × 1011) ] = 1.3 × 10-4 m Elongation of the steel wire = 1.49 × 10–4 m Elongation of the brass wire = 1.3 × 10–4 m. 9.6. The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall.A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face? Ans. Edge of the aluminium cube, L = 10 cm = 0.1 m The mass attached to the cube, m = 100 kg Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa Shear modulus, η = Shear stress / Shear strain = (F/A) / (L/ L) Where, F = Applied force = mg = 100 × 9.8 = 980 N A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m2 L = Vertical deflection of the cube ∴ L = FL / Aη= 980 × 0.1 / [ 10-2 × (25 × 109) ]= 3.92 × 10–7 m The vertical deflection of this face of the cube is 3.92 ×10–7 m. 9.7. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. Ans.Mass of the big structure, M = 50,000 kg Inner radius of the column, r = 30 cm = 0.3 m Outer radius of the column, R = 60 cm = 0.6 m Young’s modulus of steel, Y = 2 × 1011 Pa Total force exerted, F = Mg = 50000 × 9.8 N Stress = Force exerted on a single column = 50000 × 9.8 / 4 = 122500 N Young’s modulus, Y = Stress / Strain Strain = (F/A) / Y Where, Area, A = π (R2 – r2) = π ((0.6)2 – (0.3)2) Strain = 122500 / [ π ((0.6)2 – (0.3)2) × 2 × 1011 ] = 7.22 × 10-7 Hence, the compressional strain of each column is 7.22 × 10–7. 9.8. A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? Ans. Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m Area of the copper piece: A=l×b= 19.1 × 10–3 × 15.2 × 10–3= 2.9 × 10–4 m2 Tension force applied on the piece of copper, F = 44500 N
Modulus of elasticity of copper, η = 42 × 109 N/m2 Modulus of elasticity, η = Stress / Strain =(F/A) / Strain ∴ Strain = F / Aη= 44500 / (2.9 × 10-4 × 42 × 109)= 3.65 × 10–3. Q. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N m–2, what is the maximum load the cable can support? Ans. Radius of the steel cable, r = 1.5 cm = 0.015 m Maximum allowable stress = 108 N m–2 Maximum stress = Maximum force / Area of cross-section ∴ Maximum force = Maximum stress × Area of c= 108 × π (0.015)2= 7.065 × 104 N Hence, the cable can support the maximum load of 7.065 × 104 N. 9.10. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension. Answer The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since th same. The relation for Young’s modulus is given as: Y = Stress / Strain = (F/A) / Strain = (4F/πd2) / Strain ....(i) Where, F = Tension force A = Area of cross-section d = Diameter of the wire It can be inferred from equation (i) that Y ∝ (1/d2) Young’s modulus for iron, Y1 = 190 × 109 Pa Diameter of the iron wire = d1 Young’s modulus for copper, Y2 = 120 × 109 Pa Diameter of the copper wire = d2 Therefore, the ratio of their diameters is given as: 9.11. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path. Ans. Mass, m = 14.5 kg Length of the steel wire, l = 1.0 m Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2 Let l be the elongation of the wire
when the mass is at the lowest point of its path. When the mass is placed at the position of the vertical circle, the total force on the mass is: F = mg + mlω2= 14.5 × 9.8 + 14.5 × 1 × (12.56)2= 2429.53 N Young's modulus = Strss / Strain Y = (F/A) / (∆l/l) ∴ ∆l = Fl / AY Young’s modulus for steel = 2 × 1011 Pa ∆l = 2429.53 × 1 / (0.065 × 10-4 × 2 × 1011) = 1.87 × 10-3 m Hence, the elongation of the wire is 1.87 × 10–3 m. 9.12. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large. Ans. Initial volume, V1 = 100.0l = 100.0 × 10 –3 m3 Final volume, V2 = 100.5 l = 100.5 ×10 –3 m3 Increase in volume,V = V2 – V1 = 0.5 × 10–3 m3 Increase in pressure, p = 100.0 atm = 100 × 1.013 × 105 Pa Bulk modulus = p / ( V/V1) = p × V 1 / V= 100 × 1.013 × 105 × 100 × 10-3 / (0.5 × 10-3)= 2.026 × 109 Pa Bulk modulus of air = 1 × 105 Pa ∴ Bulk modulus of water / Bulk modulus of air = 2.026 × 109 / (1 × 105) = 2.026 × 104 This ratio is very high because air is more compressible than water.9.13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?Ans.Let the given depth be h.Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105 Pa Density of water at the surface, ρ1 = 1.03 × 103 kg m–3Let ρ2 be the density of water at the depth h.Let V1 be the volume of water of mass m at the surface. Let V2 be the volume of water of mass m at thedepth h. Let V be the change in volume.V = V[ 1(1-/ρV12) - (1/ρ2) ]= m∴ Volumetric strain = V / V1= m [ (1/ρ1) - (1/ρ2) ] × (ρ1 / m) V / V1 = 1 - (ρ1/ρ2) ......(i)Bulk modulus, B = pV1 / V V / V1 = p / B 10-11 Pa-1Compressibility of water = (1/B) = 45.8 ×∴ V / V1 = 80 × 1.013 × 105 × 45.8 × 10-11 = 3.71 × 10-3 ....(ii) For equations (i) and (ii), we get: 1 - (ρ1/ρ2) = 3.71 × 10-3 ρ2 = 1.03 × 103 / [ 1 - (3.71 × 10-3) ] = 1.034 × 103 kg m-3Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.Q. Compute the fractional change in volume of a glass slab, when subjected to a hydraulicpressure of 10 atm.
Ans.Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 ×1.013 × 105 Pa glass, B = 37 × 109 Nm–2 Bulk modulus, B = p / (∆V/V)Bulk modulus ofWhere, ∆V/V = Fractional change in volume ∴ ∆V/V = p / B= 10 × 1.013 × 105 / (37 × 109)= 2.73 × 10-5 Hence, the fractional change in the volume of the glass slab is 2.73 × 10–5.Q. Determine the volume contraction of a solid copper cube, 10 cm on an edge, whensubjected to a hydraulic pressure of 7.0 ×106 Pa. Answer Length of an edge of the solid copper cube, l = 10 cm = 0.1 m Hydraulic pressure, p = 7.0 × 106 Pa Bulk modulus of copper, B = 140 × 109 Pa Bulk modulus, B = p / (∆V/V) Where,∆V/V = Volumetric strain V = Change in volume V = Original volume. V = pV / B Original volume of the cube, V = l3 ∴ V = pl3 / B= 7 × 106 × (0.1)3 / (140 × 109)= 5 × 10-8 m3 = 5 × 10-2 cm-3 Therefore, the volume contraction of the solid copper cube is 5 × 10–2 cm–3. 9.16. How much should the pressure on a litre of water be changed to compress it by 0.10%? Ans. Volume of water, V = 1 L It is given that water is to be compressed by 0.10%. ∴ Fractional change, ∆V / V = 0.1 / (100 × 1) = 10-3 Bulk modulus, B = ρ / (∆V/V) ρ = B × (∆V/V) Bulk modulus of water, B = 2.2 × 109 Nm-2 ρ = 2.2 × 109 × 10-3 = 2.2 × 106 Nm-2 Therefore, the pressure on water should be 2.2 ×106 Nm–2. 9.17. Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?
AnswerDiameter of the cones at the narrow ends, d = 0.50 mm = 0.5 × 10–3 mRadius, r = d/2 = 0.25 × 10-3 m Compressional force, F = 50000 N Pressure at the tip of the anvil:P = Force / Area = 50000 / π(0.25 × 10-3)2 = 2.55 × 1011 PaTherefore, the pressure at the tip of the anvil is 2.55 × 1011 Pa.9.18. A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel(wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectionalareas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.AnswerCross-sectional area of wire A, a1 = 1.0 mm2 = 1.0 × 10–6 m2Cross-sectional area of wire B, a2 = 2.0 mm2 = 2.0 × 10–6 m2Young’s modulus for steel, Y1 = 2 × 1011 Nm–2Young’s modulus for aluminium, Y2 = 7.0 ×1010 Nm–2(a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.Stress in the wire = Force / Area = F / aIf the two wires have equal stresses, then:F1 / a1 = F2 / a2Where,F1 = Force exerted on the steel wireF2 = Force exerted on the aluminum wireF1 / F2 = a1 / a2 = 1 / 2 ....(i)The situation is shown in the following figure:Taking torque about the point of suspension, we have: F1y = F2 (1.05 - y)F1 / F2 = (1.05 - y) / y ......(ii)Using equations (i) and (ii), we can write: (1.05 - y) / y = 1 / 22(1.05 - y) = y y = 0.7 mIn order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m(b) Young's modulus = Stress / StrainStrain = Stress / Young's modulus = (F/a) / Y If the strain in the two wires is equal, then: (F1/a1) / Y1 = (F2/a2) / Y2F1 / F2 = a1Y1 / a2Y2a1 / a2 = 1/2With PDFmyURL a!
F1 / F2 = (1 / 2) (2 × 1011 / 7 × 1010) = 10 / 7 .......(iii)Taking torque about the point where mass m, is suspended at a distance y1 from the side where wireA attached, we get:F1y1 = F2 (1.05 – y1)F1 / F2 = (1.05 - y1) / y1 ....(iii) Using equations (iii) and (iv), we get: (1.05 - y1) / y1 = 10 / 77(1.05 - y1) = 10y1y1 = 0.432 mIn order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m9.19. A mild steel wire of length 1.0 m and cross-sectional area0.50 × 10–2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.AnsFrom the above figure,Let x be the depression at the mid point i.e. CD = x. In fig.,AC= CB = l = 0.5 m ; m = 100 g = 0.100 Kg AD= BD = (l2 + x2)1/2Increase in length, ∆l = AD + DB - AB = 2AD - AB9.20. Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearingstress on the rivet is not to exceed 6.9 × 107 Pa? Assume that each rivet is to carry one quarterof the load.With PDFmyURL anyone can convert entire websites to PDF!
Answer Diameter of the metal strip, d = 6.0 mm = 6.0 × 10–3 m Radius, r = d/2 = 3 × 10-3 m Maximum shearing stress = 6.9 × 107 Pa Maximum stress = Manimum load or force / Area Maximum force= 6.9 × 107 × π × (r) 2= 6.9 × 107 × π × (3 ×10–3)2= 1949.94 N Each rivet carries one quarter of the load. ∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76 N. Q. The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 ×108 Pa. A steel ball of initial volume 0.32 m3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom? Ans. Water pressure at the bottom, p = 1.1 × 108 Pa Initial volume of the steel ball, V = 0.32 m3 Bulk modulus of steel, B = 1.6 × 1011 Nm–2 The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface. Let the change in the volume of the ball on reaching the bottom of the trench be V. Bulk modulus, B = p / (∆V/V) ∆V = B / pV = 1.1 × 108 × 0.32 / (1.6 × 1011 ) = 2.2 × 10-4 m3 Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10–4 m3. 10.MECHANICAL PROPERTIES OF FLUIDS 10.1. Explain why(a) The blood pressure in humans is greater at the feet than at the brain(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of itsvalue at the sea level, though the height of the atmosphere is more than 100 km(c) Hydrostatic pressure is a scalar quantity even though pressure is force dividedby area.Answer(a) The height of the blood column in the human body is more at feet than at the brain. That is why, the blood exerts more pressure at the feet than at the brain.(Pressure = h ρ g where h = height, ρ = density of liquid and g = accleration due to gravity)(e) Density of air is the maximum near the sea level. Density of air decreases with increase in height from the surface. Ata height of about 6 km, density decreases to nearly half of its value at the sea level. Atmospheric pressure is proportionalto density. Hence, at a height of 6 km from the surface, it decreases to nearly half of its value at the sea level.
(f) When force is applied on a liquid, the pressure in the liquid is transmitted in all directions. Hence,hydrostatic pressure does not have a fixed direction and it is a scalar physical quantity.10.2. Explain why= The angle of contact of mercury with glass is obtuse, while that of water withglass is acute.= Water on a clean glass surface tends to spread out while mercury on the samesurface tends to form drops. (Put differently, water wets glass while mercury doesnot.)= Surface tension of a liquid is independent of the area of the surface= Water with detergent dissolved in it should have small angles of contact.= A drop of liquid under no external forces is always spherical in shapeAnswer(a) The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid iscalled the angle of contact (θ), as shown in the given figureSla, Ssa, and Ssl are the respective interfacial tensions betweenthe liquid-air, solid-air, and solid-liquid interfaces. At the line of contact, the surface forces between the threemedia must be in equilibrium, i.e.,Cosθ = (Ssa - Sla) / SlaThe angle of contact θ , is obtuse if Ssa < Sla (as in the case of mercury on glass). This angle is acute if Ssl <Sla (as in the case of water on glass).= Mercury molecules (which make an obtuse angle with glass) have a strong force of attraction betweenthemselves and a weak force of attraction toward solids. Hence, they tend to form drops.On the other hand, water molecules make acute angles with glass. They have a weak force of attractionbetween themselves and a strong force of attraction toward solids. Hence, they tend to spread out.= Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface.This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of theliquid surface.
= Water with detergent dissolved in it has small angles of contact (θ). This is because for a small θ, there is afast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine ofthe angle of contact (θ). If θ is small, then cosθ will be large and the rise of the detergent water in the cloth willbe fast.= A liquid tends to acquire the minimum surface area because of the presence of surface tension. The surface area ofa sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.10.3. Fill in the blanks using the word(s) from the list appended with each statement:= Surface tension of liquids generally . . . with temperatures (increases / decreases)► decreases= Viscosity of gases. .. with temperature, whereas viscosity of liquids . . . with temperature (increases /decreases)► increases ; decreases= For solids with elastic modulus of rigidity, the shearing force is proportional to . . . , while for fluids it is proportional to. .. (shear strain / rate of shear strain) ► shear strain ; rate of shear strain= For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass /Bernoulli’s principle)► conservation of mass ; Bernoulli’s principle(e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for turbulence for anactual plane (greater / smaller)► greater10.4. Explain whyA.To keep a piece of paper horizontal, you should blow over, not under, itB.When we try to close a water tap with our fingers, fast jets of water gush throughthe openings between our fingersC.The size of the needle of a syringe controls flow rate better than the thumbpressure exerted by a doctor while administering an injectionD.A fluid flowing out of a small hole in a vessel results in a backward thrust on thevesselE.A spinning. cricket ball in air does not follow a parabolic trajectoryAnswerA.When we blow over the paper, the velocity of air blow increases and hence pressure of air on it decreases (according toBeroulli's Theorem), where as pressure of air blow the paper is atmospheric. Hence, the paper stays horizontal.B.By doing so the area of outlet of water jet is reduced, so velocity of water increases according to equation ofcontinuity, Area × Velocity = Constant.
C.The small opening of a syringe needle controls the velocity of the blood flowing out. This is because of theequation of continuity. At the constriction point of the syringe system, the flow rate suddenly increases to a highvalue for a constant thumb pressure applied. D.When a fluid flows out from a small hole in a vessel, the vessel receives a backward thrust. A fluid flowing out from a small hole has a large velocity according to the equation of continuity:Area × Velocity = ConstantAccording to the law of conservation of momentum, the vessel attains a backward velocity because there are noexternal forces acting on the system.E.A spinning cricket ball has two simultaneous motions – rotatory and linear. These two types of motion oppose theeffect of each other. This decreases the velocity of air flowing below the ball. Hence, the pressure on the upper sideof the ball becomes lesser than that on the lower side. An upward force acts upon the ball. Therefore, the ball takesa curved path. It does not follow a parabolic path.10.5. A 50 kg girl wearing high heel shoes balances on a single heel. The heel iscircular with a diameter 1.0 cm. What is the pressure exerted by the heel on thehorizontal floor?AnswerMass of the girl, m = 50 kgDiameter of the heel, d = 1 cm = 0.01 mRadius of the heel, r = d/2 = 0.005 mArea of the heel = πr2= π (0.005)2–5 2= 7.85 × 10 mForce exerted by the heel on the floor:F = mg= 50 × 9.8= 490 NPressure exerted by the heel on the floor:P = Force / Area -5 6 -2= 490 / (7.85 × 10 ) = 6.24 × 10 NmTherefore, the pressure exerted by the heel on the horizontal floor is 6 –26.24 × 10 Nm .10.6. Toricelli’s barometer used mercury. Pascal duplicated itusing French wine of density 984 kg m–3. Determine the height of the wine column fornormal atmospheric pressure.AnswerDensity of mercury, ρ1 = 13.6 × 3 3 Height of the mercury column, h1 = 0.76 m Density of French wine, ρ2 = 984 3 10 kg/m kg/mHeight of the French wine column = h2
2Acceleration due to gravity, g = 9.8 m/sThe pressure in both the columns is equal, i.e.,Pressure in the mercury column = Pressure in the French wine columnρ1h1g = ρ2h2g h2 = ρ1h1 / ρ2 3= 13.6 × 10 × 0.76 / 984 = 10.5 mHence, the height of the French wine column for normal atmospheric pressure is 10.5 m.10.7. A vertical off-shore structure is built to withstand amaximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil wellin the ocean? Take the depth of ocean to be roughly 3 km, and ignore oceancurrents.Ans. 93The maximum allowable stress for the structure, P = 10 Pa Depth of the ocean, d = 3 km = 3 × 10 mDensity of water, ρ = 3 3 2 10 kg/mAcceleration due to gravity, g = 9.8 m/sThe pressure exerted because of the sea water at depth, d = ρdg 33= 3 × 10 × 10 × 9.8 7= 2.94 × 10 Pa 9The maximum allowable stress for the structure (10 Pa) is 7greater than the pressure of the sea water (2.94 × 10 Pa). The pressure exerted by the ocean is less than thepressure that the structure can withstand. Hence, the structure is suitable for putting up on top of an oil well in theocean.10.8. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area ofcross-section of the 2piston carrying the load is 425 cm . What maximum pressure would the smaller piston have to bear?AnswerThe maximum mass of a car that can be lifted, m = 3000 kg 2 –4 2Area of cross-section of the load-carrying piston, A = 425 cm = 425 × 10 mThe maximum force exerted by the load, F = mg= 3000 × 9.8= 29400 NThe maximum pressure exerted on the load-carrying -4piston, P = F/A 29400 / (425 × 10 ) 5= 6.917 × 10 PaPressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston 5would have to bear is 6.917 × 10 Pa.10.9. A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the twoarms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specificgravity of spirit?
AnswerHeight of the spirit column, h1 = 12.5 cm = 0.125 m Height of the water column, h2 = 10 cm = 0.1 m P0 =Atmospheric pressureρ1 = Density of spirit ρ2 = Density of waterPressure at point B = P0 + ρ1h1g Pressure at point D = P0 + ρ2h2gPressure at points B and D is the same. P0 + ρ1h1g = P0 + ρ2h2gρ1 / ρ2 = h2 / h1 = 10 / 12.5 = 0.8Therefore, the specific gravity of spirit is 0.8.10.10. In problem 10.9, if 15.0 cm of water and spirit each are further poured into the respective arms ofthe tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury =13.Height of the water column, h1 = 10 + 15 = 25 cm Height of the spirit column, h2 = 12.5 + 15 = 27.5 cm Density –3of water, ρ1 = 1 g cmDensity of spirit, ρ2 = 0.8 g –3 cm –3Density of mercury = 13.6 g cmLet h be the difference between the levels of mercury in the two arms.Pressure exerted by height h, of the mercury column:= hρg= h × 13.6g … (i)Difference between the pressures exerted by water and spirit:= ρ1h1g - ρ2h2g= g(25 × 1 – 27.5 × 0.8)= 3g … (ii)Equating equations (i) and (ii), we get: 13.6 hg = 3gh = 0.220588 ≈ 0.221 cmHence, the difference between the levels of mercury in the two arms is 0.221 cm.Q.Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.AnswerBernoulli’s equation cannot be used to describe the flow of water through a rapid in a river because of theturbulent flow of water. This principle can only be applied to a streamline flow.Q.Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation?Explain.Answer
No, it does not matter if one uses gauge pressure instead of absolute pressure while applying Bernoulli’sequation. The two points where Bernoulli’s equation is applied should have significantly different atmosphericpressures.10.13. Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount ofglycerine –3 –1collected per second at one end is 4.0 × 10 kg s , what is the pressure difference between the two ends of the tube?(Density 3 –3of glycerine = 1.3 × 10 kg m and viscosity of glycerine = 0.83 Pa s). [You may also like to check if theassumption of laminar flow in the tube is correct].AnswerLength of the horizontal tube, l = 1.5 m Radius of the tube, r = 1 cm = 0.01 m Diameter of the tube, d = 2r = 0.02 m –3 –1 –3 –1Glycerine is flowing at a rate of 4.0 × 10 kg s . M = 4.0 × 10 kg s –3Density of glycerine, ρ = 1.3 × 3 kg m Viscosity of glycerine, η = 0.83 Pa s Volume of glycerine flowing per sec: 10V=M/ρ -3 3 -6 3 -1= 4 × 10 / (1.3 × 10 ) = 3.08 × 10 m sAccording to Poiseville’s formula, we have the relation for the rate of flow:V = πpr4 / 8ηlWhere, p is the pressure difference between the two ends of the tube∴ p = V8ηl / πr4= 3.08 × -6 × 8 × 0.83 × 1.5 / [ π × (0.01)4 ] 10 2= 9.8 × 10 PaReynolds’ number is given by the relation:R = 4pV / πdη= 4 × 1.3 × 3 × 3.08 × -6 / ( π × 0.02 × 0.83) 10 10= 0.3Reynolds’ number is about 0.3. Hence, the flow is laminar.10.14. In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lowersurfaces of the –1 –1wing are 70 m s and 63 m s respectively. What is the lift on 2 –3the wing if its area is 2.5 m ? Take the density of air to be 1.3 kg m .AnswerSpeed of wind on the upper surface of the wing, V1 = 70 m/s Speed of wind on the lower surface of the wing, V2 2= 63 m/s Area of the wing, A = 2.5 mDensity of air, ρ = 1.3 kg –3 mAccording to Bernoulli’s theorem, we have the relation: P1 + (1/2)ρV12 = P2 + (1/2)ρV22 2 2P2 - P1 = (1/2) ρ (V1 - V2 ) Where,P1 = Pressure on the upper surface of the wing P2 = Pressure on the lower surface of the wing
The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.Lift on the wing = (P2 - P1) ASave web pages as PDF manually or automatically with PDFmyURL
= (1/2) ρ 2 - 2 A (V1 V2 ) 22= (1/2) × 1.3 × [ 70 - 633 ] × 2.5= 1512.87 N = 1.51 × 10 N 3Therefore, the lift on the wing of the aeroplane is 1.51 × 10 N.10.15. Figures 10.23 (a) and (b) refer to the steady flow of a (non-viscous) liquid.Which of the two figures is incorrect? Why?AnswerFig. (a) is incorrect. Accoridng to equation of continuity, i.e., av = Constant, where area of cross-section of tube is less,the velcoity of liquid flow is more. So the velocity of liquid flow at a constriction of tube is more than the other portion of tube.Accroding to Bernoulli's Theorem, P + 1/2ρv2 = Constant, where v is more, P is less and vice versa. 210.16. The cylindrical tube of a spray pump has a cross-section of 8.0 cm one end of which has 40 fine holeseach of diameter –11.0 mm. If the liquid flow inside the tube is 1.5 m min , what is the speed of ejection of the liquid throughthe holes?Answer 2– Area of cross-section of the spray pump, A1 = 8 cm = 8 × 10 4 m2Number of holes, n = 40 –3 –3Diameter of each hole, d = 1 mm = 1 × 10 m Radius (o0f.e5a×ch1h0o–le3, )r2=md/22 = 0.5 × 10 mArea of cross-section of each hole, a = πr2 = π Total area of 40 holes, A2 = n × a 10–3)2= 40 × π (0.5 × 2 2 –6 m= 31.41 × 10 mSpeed of flow of liquid inside the tube, V1 = 1.5 m/min = 0.025m/sSpeed of ejection of liquid through the holes = V2 According to the law of continuity, we have:A1V1 = A2V2V2 = A1V1 / A2 -4 -6= 8 × 10 × 0.025 / (31.61 × 10 )= 0.633 m/sTherefore, the speed of ejection of the liquid through the holes is 0.633 m/s.
10.17. A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wireand the –2light slider supports a weight of 1.5 × 10 N (which includes the small weight of the slider). The lengthof the slider is 30 cm. What is the surface tension of the film?Ans. –2The weight that the soap film supports, W = 1.5 × 10 N Length of the slider, l = 30 cm = 0.3 mA soap film has two free surfaces. ∴ Total length = 2l = 2 × 0.3 = 0.6 m Surface tension, S = Force or Weight / 2l -2= 1.5 × 10 / 0.6 -2= 2.5 × 10 N/m –2 –1Therefore, the surface tension of the film is 2.5 × 10 N m .10.18. Figure 10.24 (a) shows a thin liquid film supporting a –2small weight = 4.5 × 10 N. What is the weight supported by a film of the same liquid at the sametemperature in Fig. (b) and (c)? Explain your answer physically.AnswerTake case (a):The length of the liquid film supported by the weight, l = 40 cm = 0.4 cm –2The weight supported by the film, W = 4.5 × 10 N A liquid film has two free surfaces.∴ Surface tension = W / 2l -2 -2= 4.5 × 10 / (2 × 0.4) = 5.625 × 10 N/mIn all the three figures, the liquid is the same. Temperature is also the same for each case. Hence, the surface –2 –1tension in figure (b) and figure (c) is the same as in figure (a), i.e., 5.625 × 10 N m .Since the length of the –2film in all the cases is 40 cm, the weight supported in each case is 4.5 × 10 N.10.19. What is the pressure inside the drop of mercury of radius 3.00 mm at roomtemperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10–1 Nm–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure insidethe drop.Ans.
–3 –1 –1Radius of the mercury drop, r = 3.00 mm = 3 × 10 m Surface tension of mercury, S = 4.65 × 10 N m Atmospheric 5pressure, P0 = 1.01 × 10 PaTotal pressure inside the mercury drop= Excess pressure inside mercury + Atmospheric pressure= 2S / r + P0 -1 -3 5= [ 2 × 4.65 × 10 / (3 × 10 ) ] + 1.01 × 10= 1.0131 × 510 5= 1.01 × 10 PaExcess pressure = 2S / r -1 -3= [ 2 × 4.65 × 10 / (3 × 10 ) ] = 310 Pa.10.20. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surfacetension of –2 –1soap solution at the temperature (20 °C) is 2.50 × 10 N m ? If an air bubble of the same dimension were formed atdepth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressureinside the 5 bubble? (1 atmospheric pressure is 1.01 × 10 Pa).AnswerExcess pressure inside the soap bubble is 20 Pa; 5 –3Pressure inside the air bubble is 1.06 × 10 Pa Soap bubb–le2is of r–ad1ius, r = 5.00 mm = 5 × 10 mSurface tension of the soap solution, S = 2.50 × 10 Nm Relative density of the soap solution = 1.20∴ Density of the soap solution, ρ = 1.2 × 3 3 Air bubble formed at a depth, h = 40 cm = 0.4 m 10 kg/m –3 5Radius of the air bubble, r = 5 mm = 5 × 10 m 1 atmospheric pressure = 1.01 × 10 Pa 2Acceleration due to gravity, g = 9.8 m/sHence, the excess pressure inside the soap bubble is given by the relation:P = 4S / r -2 -3= 4 × 2.5 × 10 / 5 × 10= 20 PaTherefore, the excess pressure inside the soap bubble is 20 Pa. The excess pressure inside the air bubble isgiven by the relation:P' = 2S / r -2 -3= 2 × 2.5 × 10 / (5 × 10 )= 10 PaTherefore, the excess pressure inside the air bubble is 10 Pa. At a depth of 0.4 m, the total pressure inside the air bubble= Atmospheric pressure + hρg + P’ 53= 1.01 × 10 + 0.4 × 1.2 × 10 × 9.8 + 10 5= 1.06 × 10 Pa 5Therefore, the pressure inside the air bubble is 1.06 × 10 P 210.21. A tank with a square base of area 1.0 m is divided by a vertical partition in the middle. The bottom of thepartition has
2a small-hinged door of area 20 cm . The tank is filled with water in one compartment, and an acid (ofrelative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep thedoor close.AnswerBase area of the given tank, A = 1.0 m 2Area of the hinged door, a = 20 2 = 20 × –4 2 Density of water, ρ1 = 3 3 cm 10 m 10 kg/mDensity of acid, ρ2 = 1.7 × 3 3 Height of the water column, h1 = 4 m Height of the acid column, h2 = 4 m 10 kg/mAcceleration due to gravity, g = 9.8 Pressure due to water is given as:P1 = h1ρ1g 34= 4 × 10 × 9.8 = 3.92 × 10 Pa Pressure due to acid is given as:P2 = h2ρ2g 34= 4 × 1.7 × 10 × 9.8 = 6.664 × 10 PaPressure difference between the water and acid columns:P = P2 - P1 44= 6.664 × 10 - 3.92 × 10 4= 2.744 × 10 PaHence, the force exerted on the door = P × a 4 –4= 2.744 × 10 × 20 × 10= 54.88 N Therefore, the force necessary to keep the door closed is 54.88 N.10.22. A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When apump removes some of the gas, the manometer reads as in Fig. 10.25 (b) The liquid used in themanometers is mercury and the atmospheric pressure is 76 cm of mercury.(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm ofmercury.(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are pouredinto the right limb of the manometer? (Ignore the small change in the volume of the gas).Answer(a) For figure (a)Atmospheric pressure, P0 = 76 cm of Hg
Difference between the levels of mercury in the two limbs gives gauge pressureHence, gauge pressure is 20 cm of Hg.Absolute pressure = Atmospheric pressure + Gauge pressure = 76 + 20 = 96 cm of Hg For figure (b)Difference between the levels of mercury in the two limbs = –18 cmHence, gauge pressure is –18 cm of Hg.Absolute pressure = Atmospheric pressure + Gauge pressure = 76 cm – 18 cm = 58 cm(b) 13.6 cm of water is poured into the right limb of figure (b). Relative density of mercury = 13.6Hence, a column of 13.6 cm of water is equivalent to 1 cm of mercury.Let h be the difference between the levels of mercury in the two limbs.The pressure in the right limb is given as: PR = Atmospheric pressure + 1 cm of Hg= 76 + 1 = 77 cm of Hg … (i)The mercury column will rise in the left limb.Hence, pressure in the left limb, PL = 58 + h .....(ii)Equating equations (i) and (ii), we get: 77 = 58 + h∴h = 19 cmHence, the difference between the levels of mercury in the two limbs will be 19 cm.10.23. Two vessels have the same base area but different shapes. The first vessel takes twice thevolume of water that the second vessel requires to fill upto a particular common height. Is the forceexerted by the water on the base of the vessel the same in the two cases? If so, why do the vesselsfilled with water to that same height give different readings on a weighing scale?Two vessels having the same base area have identical force and equal pressure acting on theircommon base area. Since the shapes of the two vessels are different, the force exerted on the sides ofthe vessels has non-zero vertical components. When these vertical components are added, the totalforce on one vessel comes out to be greater than that on the other vessel. Hence, when these vesselsare filled with water to the same height, they give different readings on a weighing scale.10.24. During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At whatheight must the blood container be placed so that blood may just enter the vein? [Use the density of whole bloodfrom Table 10.1].AnswerGauge pressure, P = 2000 Pa –3 mDensity of whole blood, ρ = 1.06 × 3 2kg 10Acceleration due to gravity, g = 9.8 m/s Height of the blood container = h Pressure of the blood container, P =hρg ∴ h = P / ρg 3= 2000 / (1.06 × 10 × 9.8)= 0.1925 mThe blood may enter the vein if the blood container is kept at a height greater than 0.1925 m, i.e., about 0.2 m.
10.25. In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in thepotential and kinetic energy. (a) What is the largest average velocity of blood –3flow in an artery of diameter 2 × 10 m if the flow must remain laminar? (b) Do the dissipative forces becomemore important as the fluid velocity increases? Discuss qualitatively.Answer(a) If dissipative forces are present, hen some forces in liquid flow due to pressure difference is spent against dissipativeforces. Due to which the pressure drop becomes large.(b) The dissipative forces become more important with increasing flow velocity, because of turbulence.10.26. (a) What is the largest average velocity of blood flow in –3an artery of radius 2 × 10 m if the flow must remain laminar? –3(b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10 Pa s).Answer –3(a) Radius of the artery, r = 2 × 10 m –3 –Diameter of the artery, d = 2 × 2 × 10 m = 4 × 10 3 m Viscosity of blood, η = 2.084 X 10-3 Pa sDensity of blood, ρ = 1.06 × 3 3 Reynolds’ number for laminar flow, NR = 2000 10 kg/mThe largest average velocity of blood is given by the relation:Varg = NRη / ρd -3 3 -3= 2000 × 2.084 × 10 / (1.06 × 10 × 4 × 10 )= 0.983 m/sTherefore, the largest average velocity of blood is 0.983 m/s.(b) Flow rate is given by the relation:R = π 2 r Vavg -3 2= 3.14 × (2 × 10 ) × 0.983 -5 3 -1= 1.235 × 10 m s -5 3 -1Therefore, the corresponding flow rate is 1.235 × 10 m s .10.27. A plane is in level flight at constant speed and each of its 2two wings has an area of 25 m . If the speed of the air is 180 km/h over the lower wing and 234 km/hover the upper wing surface, determine the plane’s mass. (Take air density 1 12 –3 22 Pto be 1 kg m ). 1Answer 2The area of the wings of the plane, A = 2 × 25 = 50 m Speed of air over the lower wing, V1 = 180 km/h = 50 m/sSpeed of air over the upper wing, V2 = 234 km/h = 65 m/s –3Density of air, ρ = 1 kg mPressure of air over the lower wing = P1 Pressure of air over the upper wing= P2The upward force on the plane can be obtained using Bernoulli’s equation as:
P1 +(1/2) ρ V12=P2 + (1/2) ρ V22P1-P2=(1/2) ρ (V22-V12)………(1)The upward force (F) on the plane can be calculated as: (P1 - P2)A = (1/2) ρ ( 2 - 2 A V2 V1 ) 22= (1/2) × 1 × (65 - 50 ) × 50= 43125 NUsing Newton’s force equation, we can obtain the mass (m) of the plane as:F = mg∴ m = 43125 / 9.8 = 4400.51 kg ∼ 4400 k .Hence, the mass of the plane is about 4400 kg.10.28. In Millikan’s oil drop experiment, what is the terminal speed of an unchargeddrop of radius 2.0 × 10–5 m and density 1.2 × 103 kg m–3? Take the viscosity of air atthe temperature of –5the experiment to be 1.8 × 10 Pa s. How much is the viscous force on the drop at that speed? Neglectbuoyancy of the drop due to air.AnS. –10Terminal speed = 5.8 cm/s Viscous force = 3.9 × 10 NRadius of the given uncharged drop, r = 2.0 × –5 m Density of the uncharged drop, ρ = 1.2 × 3 kg –3 -5 10 10 m 10Viscosity of air, η = 1.8 × Pa sDensity of air (ρ0) can be taken as zero in order to neglect buoyancy of air.Acceleration due to gravity, g = 9.8 2 Terminal velocity (v) is given by the relation: v = 2 × (ρ - ρ0) g / 9η m/s 2r -5 2 3 -5= 2 × (2 × 10 ) (1.2 × 10 - 0 ) × 9.8 / (9 × 1.8 × 10 ) -2= 5.8 × 10 m/s -1= 5.8 cm s –1Hence, the terminal speed of the drop is 5.8 cm s . The viscous force on the drop is given by:F = 6πηrv -5 -5 -2 -10∴ F = 6 × 3.14 × 1.8 × 10 × 2 × 10 × 5.8 × 10 = 3.9 × 10 N –10Hence, the viscous force on the drop is 3.9 × 10 N.Q. Mercury has an angle of contact equal to 140° with soda lime glass. A narrowtube of radius 1.00 mm made of this glass is dipped in a trough containing mercury.By what amount does the mercury dip down in the tube relative to the liquid surfaceoutside? Surface tension of mercury at the temperatureof the experiment is 0.465 N m–1. Density of mercury = 13.6 × 103 kg m–3.AnswerAngle of contact between mercury and soda lime glass, θ = 140° –3Radius of the narrow tube, r = 1 mm = 1 × 10 m –1Surface tension of mercury at the given temperature, s = 0.465 N m 3 3Density of mercury, ρ =13.6 × 10 kg/m Dip in the height of mercury = h 2Acceleration due to gravity, g = 9.8 m/s
Surface tension is related with the angle of contact and the dip in the height as:s = hρgr / 2 Cosθ∴ h = 2s Cosθ / ρgr 0 -3 3= 2 × 0.465 × Cos140 / (1 × 10 × 13.6 × 10 × 9.8)= -0.00534 m= -5.34 mHere, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm.10.30. Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to forma U-tube open at both ends. If the U-tube contains water, what is the difference in itslevels in the two limbs of the tube? Surface tension of water at the temperature of theexperiment is 7.3 × 10–2 N m–1. Take the angle of contact to be zero and density ofwater to be 1.0 × 103 kg m–3 (g = 9.8 m s–2).Answer –3 -3Diameter of the first bore, d1 = 3.0 mm = 3 × 10 m Hence, the radius of the first bore, r1 = d1/ 2 = 1.5 × 10 m Diameter –3 -3of the first bore, d2 = 6.0 mm = 6 × 10 mm Hence, the radius of the first bore, r2 = d2/ 2 = 3 × 10 m –2 –1Surface tension of water, s = 7.3 × 10 N m –3Angle of contact between the bore surface and water, θ= 0 Density of water, ρ =1.0 × 3 kg/m 2 10Acceleration due to gravity, g = 9.8 m/sLet h1 and h2 be the heights to which water rises in the first andsecond tubes respectively. These heights are given by the relations:h1 = 2s Cosθ / r1ρg .....(i)h2 = 2s Cosθ / r2ρg .....(ii)The difference between the levels of water in the two limbs of the tube can be calculated as: -3= 4.966 × 10 m= 4.97 manually or automatically with PDFmyURL Hence, the difference between levels of water in the two bores is 4.97 mm.10.31. (a) It is known that density ρ of air decreases with heighty as ρ0e-y/y0Where ρ0 = 1.25 kg m-3
is the density at sea level, and y0 is a constant. This density variation is called the lawof atmospheres. Obtain this law assuming that the temperature of atmosphereremains a constant (isothermal conditions). Also assume that the value of g remainsconstant.(b) A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assumethat the balloon maintains constant radius as it rises. How high does it rise?[Take y0= 8000 m and ρHe = 0.18 kg m-3 ] 3Ans.(a) Volume of the balloon, V = 1425 m Mass of the payload, m = 400 kg 2Acceleration due to gravity, g = 9.8 m/sy0 = 8000 m -3ρHe = 0.18 kg m ....(i)ρ0 = 1.25 kg -3 mDensity of the balloon = ρHeight to which the balloon rises = yDensity (ρ) of air decreases with height (y) as:Ρ=ρ0e=y/y0Ρ/ρ0=e=y/y0 This density variation is called the law of atmospherics.It can be inferred from equation (i) that the rate of decrease of density with height is directly proportional to ρ, i.e.,-(dρ / dy) ∝ρ (dρ / dy) = -kρ(dρ / ρ) = -k dyWhere, k is the constant of proportionalityHeight changes from 0 to y, while density changes from ρ0 to ρ Integrating the sides between these limits, weget:Comparing equations (i) and (ii), we get: y0 = 1/k
k = 1/y0 ....(iii)From equations (i) and (iii), we getρ = ρ0e-y/y0(b) Density ρ = Mass / Volume= (Mass of the payload + Mass of helium) / Volume ( m + VρHe) / V= (400 + 1425 × 0.18) / 1425 -3= 0.46 kg mFrom equations (ii) and (iii), we can obtain y as: ρ = ρ0e-y/y0loge(ρ / ρ0) = -y / y0∴ y = - 8000 × loge(0.46 / 1.25)=-8000 × (-1)= 8000 m = 8 kmHence, the balloon will rise to a height of 8 km. 11.THERMAL PROPERTIES OF MATTER Q. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales. Answer Kelvin and Celsius scales are related as: TC = TK – 273.15 … (i) Celsius and Fahrenheit scales are related as: TF = (9/5)TC + 32 ....(ii) For neon: TK = 24.57 K ∴ TC = 24.57 – 273.15 = –248.58°C TF = (9/5)TC + 32 (c) (9/5) × (-248.58) +32 0 (d) 415.44 F For carbon dioxide: TK = 216.55 K ∴ TC= 216.55 – 273.15 = –56.60°C TF = (9/5)TC + 32 0 (g) (9/5) × (-56.6 ) +32 0 (g) -69.88 C Q. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB? Answer
Triple point of water on absolute scaleA, T1 = 200 A Triple point of water on absolute scale B, T2 = 350 B Triple point ofwater on Kelvin scale, TK = 273.15 KThe temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A.T1 = TK200 A = 273.15 K∴ A = 273.15/200The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.T2 = TK350 B = 273.15 ∴ B = 273.15/350TA is triple point of water on scale A. TB is triple point of water on scale B. ∴ 273.15/200 × TA = 273.15/350 × TBTherefore, the ratio TA : TB is given as 4 : 7.11.3. The electrical resistance in ohms of a certain thermometer varies with temperatureaccording to the approximate law:R = Ro [1 + α (T – To)]The resistance is 101.6 Ω at the triple-point of water 273.16 K,and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when theresistance is 123.4 Ω?AnswerIt is given that:R = R0 [1 + α (T – T0)] … (i) where,R0 and T0 are the initial resistance and temperature respectivelyR and T are the final resistance and temperature respectively α is a constantAt the triple point of water, T0 = 273.15 K Resistance of lead, R0 = 101.6 ΩAt normal melting point of lead, T = 600.5 K Resistance of lead, R = 165.5 ΩSubstituting these values in equation (i), we get: R = Ro [1 + α (T – To)]165.5 = 101.6 [ 1 + α(600.5 - 273.15) ] 1.629 = 1 + α (327.35)∴ α = 0.629 / 327.35 = 1.92 × -3 -1 10 KFor resistance, R1 = 123.4 ΩR1 = R0 [1 + α (T – T0)] where,T is the temperature when the resistance of lead is 123.4 Ω -3123.4 = 101.6 [ 1 + 1.92 × 10 ( T - 273.15) ] Solving for T, we getT = 384.61 K.11.4. Answer the following:= The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong intaking the melting point of ice and the boiling point of water as standard fixed points (as was originallydone in the Celsius scale)?= There were two fixed points in the original Celsius scale as mentioned above which were assigned thenumber 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point ofwater, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed pointon this (Kelvin) scale?
= The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale bytc = T – 273.15Why do we have 273.15 in this relation, and not 273.16?(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size isequal to that of the Fahrenheit scale?Answer= The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triplepoint of water is always 273.16 K. The melting point of ice and boiling point of water do not have particular valuesbecause these points depend on pressure and temperature.= The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.= The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature 0°Con Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K.Hence, absolute temperature (Kelvin scale) T, is related to temperature tc, on Celsius scale as:tc = T – 273.15(d) Let TF be the temperature on Fahrenheit scale and TK be thetemperature on absolute scale. Both the temperatures can be related as:(TF - 32) / 180 = (TK - 273.15) / 100 ....(i)Let TF1 be the temperature on Fahrenheit scale and TK1 be thetemperature on absolute scale. Both the temperatures can be related as:(TF1 - 32) / 180 = (TK1 - 273.15) / 100 ....(ii)It is given that: TK1 – TK = 1 KSubtracting equation (i) from equation (ii), we get: (TF1 - TF) / 180 = (TK1 - T K) / 100 = 1 / 100TF1 - TF = (1 ×180) / 100 = 9/5 Triple point of water = 273.16 K∴ Triple point of water on absolute scale = 273.16 × (9/5) = 491.6911.5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The followingobservations are made:Temperature Pressure PressureTriple-point of thermometer A thermometer Bwater 1.250 ×105 Pa 0.200 ×105 PaNormal melting 1.797 ×105 Pa 0.287 ×105 Papoint of sulphur
= What is the absolute temperature of normal melting point of sulphur as read by thermometersAand B?= What do you think is the reason behind the slight difference in answers of thermometers Aand B? (Thethermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancybetween the two readings?Answer(a) Triple point of water, T = 273.16 K. 5At this temperature, pressure in thermometer A, PA = 1.250 × 10PaLet T1 be the normal melting point of sulphur. 5At this temperature, pressure in thermometer A, P1 = 1.797 × 10PaAccording to Charles’ law, we have the relation:PA / T = P1 / T1 5 10 )∴ T1 = (P1T) / PA = (1.797 × 10 × 273.16) / (1.250 ×= 392.69 KTherefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K.At triple point 273.16 K, the pressure in thermometer B, PB = 50.200 × 10 PaAt temperature T1, the pressure in thermometer B, P2 = 0.287 × 510 PaAccording to Charles’ law, we can write the relation:PB / T = P1 / T1 55(0.200 × 10 ) / 273.16 = (0.287 × 10 ) / T1PDFmyURL lets you convert a complete website to PDF automatically! 55∴ T1 = [ (0.287 × 10 ) / (0.200 × 10 ) ] × 273.16 = 391.98 KTherefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.(b) The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect idealgases. Hence, there is a slight difference between the readings of thermometers A and B.To reduce the discrepancy between the two readings, the experiment should be carried under low pressureconditions. At low pressure, these gases behave as perfect ideal gases.11.6. A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rodmeasured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is theactual length of the steel rod on that day? What is the length of the same steel rod on a day when thetemperature is 27.0 °C? Coefficient of linear –5 –1expansion of steel = 1.20 × 10 K .Ans.Length of the steel tape at temperature T = 27°C, l = 1 m = 100 cmAt temperature T1 = 45°C, the length of the steel rod, l1 = 63 cm
Coefficient of linear expansion of steel, α = 1.20 × –5 –1 10 KLet l2 be the actual length of the steel rod and l' be the length ofthe steel tape at 45°C. l' =l+ αl(T1 - T) -5∴ l' = 100 + 1.20 × 10 × 100(45 - 27) = 100.0216 cmHence, the actual length of the steel rod measured by the steel tape at 45°C can be calculatedas:automatically!l2 = (100.0216 / 100) × 63 = 63.0136 cmTherefore, the actual length of the rod at 45.0°C is 63.0136 cm. Its length at 27.0°C is 63.0 cm.11.7. A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of theshaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dryice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linearexpansion of the steel to be constant over the requiredtemperature range: α steel = 1.20 × –5 –1 10 K.Ans.The given temperature, T = 27°C can be written in Kelvin as: 27 + 273 = 300 KOuter diameter of the steel shaft at T, d1 = 8.70 cm Diameter of the central hole in the wheel at T, d2 = 8.69 cmCoefficient of linear expansion of steel, αsteel = –5 –1 1.20 × 10 KAfter the shaft is cooled using ‘dry ice’, its temperature becomes T1.The wheel will slip on the shaft, if the change in diameter, d = 8.69 – 8.70= – 0.01 cmTemperature T1, can be calculated from the relation: d= d1αsteel (T1 – T) (T1 – 300) (T1 – 300) = 95.780.01 = 8.70 × 1.20 × –5 10∴ T1= 204.21 K= 204.21 – 273.16= –68.95°CTherefore, the wheel will slip on the shaft when the temperature of the shaft is –69°C.11.8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change inthe diameter of the hole when the sheet is heated to 227 °C? Coefficient of –5 –1linear expansion of copper = 1.70 × 10 K .AnswerInitial temperature, T1 = 27.0°C Diameter of the hole at T1, d1 = 4.24 cm Final temperature, T2 = 227°CDiameter of the hole at T2 = d2 –5 –1 10 KCo-efficient of linear expansion of copper, αCu= 1.70 ×For co-efficient of superficial expansion β,and change in temperature T, we have the relation:Change in area (∆) / Original area (A) = β∆T [ (πd22/ 4) - (πd12 / 4) ] / (πd11 / 4) = ∆A / A= ∆A / A = 2 - 2 / 2 β = 2α (d2 d1 ) d1 But
= 2 2 / 2 = 2α∆T (d2 - d1 ) d1 22(d2 / d1 ) - 1 = 2α(T2 - T1) 22 -5 2d2 / 4.24 = 2 × 1.7 × 10 (227 - 27) +1 d2 = 17.98 × 1.0068 = 18.1∴ d2 = 4.2544 cm d2 – d1 = 4.2544 – 4.24 = 0.0144 cm Hence, the diameter increases by 1.44 × –2 cm.nvert a in diameter = 10Changecomplete website to PDF automatically!11.9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire iscooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm?Co-efficient of linear expansion –5 K–1;of brass = 2.0 × 10 Young’s modulus of brass = 0.91 × 1011 Pa.AnswerInitial temperature, T1 = 27°CLength of the brass wire at T1, l = 1.8 m Final temperature, T2 = –39°C –3Diameter of the wire, d = 2.0 mm = 2 × 10 m Tension developed in the wire = F –5 –1Coefficient of linear expansion of brass, α = 2.0 × 10 KYoung’s modulus of brass, Y = 0.91 × 11 Pa Young’s modulus is given by the relation: γ = Stress / Strain = (F/A) / 10(∆L/L)∆L = F X L / (A X Y) ......(i)Where,F = Tension developed in the wire= = Area of cross-section of the wire. L = Change in the length, given by the relation: L = αL(T2 – T1) … (ii)Equating equations (i) and (ii), we get: αL(T2 - T1) = FL / [ π(d/2)2 X Y ]= = α(T2 - T1)πY(d/2)2 2 -5 11 -3 2== 2 × 10 × (-39-27) × 3.14 × 0.91 × 10 × (2 × 10 / 2 ) = -3.8 × 10 N(The negative sign indicates that the tension is directedinward.) 2Hence, the tension developed in the wire is 3.8 ×10 N.11.10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length anddiameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to –5 – 1 –5 –1expand (Co-efficient of linear expansion of brass = 2.0 × 10 K , steel = 1.2 × 10 K ).AnswerInitial temperature, T1 = 40°C Final temperature, T2 = 250°C
Change in temperature, T = T2 – T1 = 210°C Length of the brass rod at T1, l1 = 50 cm Diameter ofthe brass rod at T1, d 1 = 3.0 mm Length of the steel rod at T2, l2 = 50 cm Diameter of the steel rodat T2, d 2 = 3.0 mm –5 –1 10 KCoefficient of linear expansion of brass, α1 = 2.0 ×Coefficient of linear expansion of steel, α2 = 1.2 × –5 –1 10 KFor the expansion in the brass rod, we have:Change in length (∆l1) / Original length (l1) = α1 T -5∴ ∆l1 = 50 × (2.1 × 10 ) × 210= 0.2205 cmFor the expansion in the steel rod, we have:Change in length (∆l2) / Original length (l2) = α2 T -5∴ ∆l1 = 50 × (1.2 × 10 ) × 210= 0.126 cmPDFmyURL lets you convert a complete website to PDF automatically!Total change in the lengths of brass and steel, l = l1 + l2= 0.2205 + 0.126= 0.346 cmTotal change in the length of the combined rod = 0.346 cm Since the rod expands freely from both ends, nothermal stress is developed at the junction. –5 –111.11. The coefficient of volume expansion of glycerin is 49 ×10 K . What is the fractional change inits density for a 30 °C rise in temperature?AnswerCoefficient of volume expansion of glycerin, αV = 49 × –5 –1 10 KRise in temperature, T = 30°C Fractional change in its volume = V/VThis change is related with the change in temperature as: V/V = αV T α TV -V =VT2 T1 T1 V(m /ρT2)- (m /ρT1) = (m /ρT1)αV T Where,m = Mass of glycerine ρT1 = Initial density at T1ρT2 = Initial density at T2(ρT1 - ρT2 ) / ρT2 = Fractional change in density –5 –2∴ Fractional change in the density of glycerin = 49 ×10 × 30 = 1.47 × 10 .Q.A 10 kW drilling machine is used to drill a bore in a small aluminium block ofmass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes,assuming 50% ofpower is used up in heating the machine itself or lost to the surroundings.Specific heat of aluminium = 0.91 J g–1 K–1.
Ans. 3Power of the drilling machine, P = 10 kW = 10 × 10 W 3Mass of the aluminum block, m = 8.0 kg = 8 × 10 gTime for which the machine is used, t = 2.5 min = 2.5 × 60 = 150 s –1 –1Specific heat of aluminium, c = 0.91 J g KRise in the temperature of the block after drilling = δT Total energy of the drilling machine = Pt 3= 10 × 10 × 150 6= 1.5 × 10 JIt is given that only 50% of the power is useful. 5 6 10Useful energy, ∆Q = (50/100) × 1.5 × 10 = 7.5 × J But ∆Q = mc∆T∴ ∆T = ∆Q / mc 53= (7.5 × 10 ) / (8 × 10 × 0.91) o= 103 CTherefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.Q.A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and thenplaced on a large ice block. What is the maximum amount of ice that can melt? (Specificheat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g– 1).Ans. θMass of the copper block, m = 2.5 kg = 2500 gRise in the temperature of the copper block, θ = 500°C –1 –1Specific heat of copper, C = 0.39 J –g1 °CHeat of fusion of water, L = 335 J gThe maximum heat the copper block can lose, Q = mC= 2500 × 0.39 × 500= 487500 JLet m1 g be the amount of ice that melts when the copper blockis placed on the ice block.The heat gained by the melted ice, Q = m1L ∴ m1 = Q / L = 487500 / 335 = 1455.22 gHence, the maximum amount of ice that can melt is 1.45 kg.Q. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at150 °C is dropped in a coppercalorimeter (of water equivalent 0.025 kg)containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute thespecific heat of the metal. If heat losses to the surroundings are not negligible, isyour answer greater or smaller than the actual value for specific heat of themetal?Ans.Mass of the metal, m = 0.20 kg = 200 g Initial temperature of the metal, T1 = 150°CFinal temperature of the metal, T2 = 40°C ’Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g 3Volume of water, V = 150 cmMass (M) of water at temperature T = 27°C: 150 × 1 = 150 gFall in the temperature of the metal: T = T1 – T2 = 150 – 40 = 110°C
Specific heat of water, Cw = 4.186 J/g/°K Specific heat of the metal = CaHeat lost by the metal, θ = mC T … (i)Rise in the temperature of the water and calorimeter system: ’ T′ = 40 – 27 = 13°C ’Heat gained by the water and calorimeter system: Δθ′′ = m1 Cw T= (M + m′) Cw ’ T … (ii)Heat lost by the metal = Heat gained by the water and colorimeter system ’’mC T = (M + m ) Cw T200 × C × 110 = (150 + 25) × 4.186 × 13∴ C = (175 × 4.186 × 13) / (110 × 200) = 0.43 -1 -1 Jg KIf some heat is lost to the surroundings, then the value of C will be smaller than the actual value.Q.Given below are observations on molar specific heats at room temperature of some commongases.The measured molar specific heats of these gases are markedly different from those formonatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain thisdifference. What can you infer from the somewhat larger (than the rest) value for chlorine?The gases listed in the given table are diatomic. Besides the translational degree of freedom, they haveother degrees of freedom (modes of motion).Heat must be supplied to increase the temperature of these gases. This increases the average energy of allthe modes of motion. Hence, the molar specific heat of diatomic gases is more than that of monatomicgases.If only rotational mode of motion is considered, then the molar specific heat of a diatomic gas = (5/2)R -1 -1= (5/2) × 1.98 = 4.95 cal mol KWith the exception of chlorine, all the observations in the given table agree with (5/2)R.This is because at room temperature, chlorine also has vibrational modes of motion besides rotational andtranslational modes of motion.Q. Answer the following questions based on the P-T phase diagram of carbon dioxide:= At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist inequilibrium?= What is the effect of decrease of pressure on the fusion and boiling point of CO2?= What are the critical temperature and pressure for CO2?What is their significance?(d) Is CO2 solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60° C under 10 atm, (c) 15 °C under 56atm?Ans.The P-T phase diagram for CO2 is shown in the following figure:
= The solid, liquid and vapour phase of carbon dioxide exist in equilibrium at the triple point, i.e.,temprature = - 56.6° C and pressure = 5.11 atm.= With the decrease in pressure, both the fusion and boiling point of carbon dioxide will decrease.= For carbon dioxide, the critical temperature is 31.1° C and critical pressure is 73.0 atm. If the tempratureof carbon dioxide is more than 31.1° C, it can not be liquified, however large pressure we may apply.= Carbon dioxide will be (a) a vapour, at =70° C under 1atm.= a solid, at -6° C under 10 atm (c) a liquid, at 15° C under 56 atm.Q.Answer the following questions based on the P–T phase diagram of CO2:(a) CO2 at 1 atm pressure and temperature – 60 °C iscompressed isothermally. Does it go through a liquid phase?(b) What happens when CO2 at 4 atm pressure is cooled from room temperature atconstant pressure?(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure andtemperature –65 °C as it is heatedup to room temperature at constant pressure.(d) CO2 is heated to a temperature 70° C and compressedisothermally. What changes in its properties do you expect to observe?Ans.The P-T phase diagram for CO2 is shown in the following figure:= Since the temprature -60° C lies to the left of 56.6° C on the curve i.e. lies in the region vapour and solidphase, so carbon dioxide will condense directly into the solid without becoming liquid.= Since the pressure 4 atm is less than 5.11 atm the carbon dioxide will condense directly into solid withoutbecoming liquid.(c) When a solid CO2 at 10 atm pressure and -65° C tempratureis heated, it is first converted into liquid. A further increase in temprature brings it into the vapour phase. At P = 10 atm,if a horizontal line is drawn parallel to the T-axis, then the points of intersection of this line with the fusion andvaporization curve will give the fusion and boiling points of CO2 at 10 atm.(d) Since 70° C is higher than the critical temprature of CO2, so the CO2 gas can not be converted into liquid state onbeing
compressed isothermally at 70° C. It will remain in the vapour state. However, the gas will depart more and more fromits perfect gas behaviour with the increase in pressure.11.18. A child running a temperature of 101°F is given an antipyrin (i.e. a medicine thatlowers fever) which causes an increase in the rate of evaporation of sweat from his body. Ifthe fever is brought down to 98 °F in 20 min, what is the average rate of extra evaporationcaused, by the drug? Assume the evaporation mechanism to be the only way by which heatis lost. The mass of the child is 30 kg. The specific heat of human body is approximately thesame as that of water, and latent heat of evaporation of water at that temperature is about –1580 cal g .Ans.Initial temperature of the body of the child, T1 = 101°F Final temperature of the body of the child, T2 = 98°F oChange in temperature, T = [ (101 - 98) × (5/9) ] C Time taken to reduce the temperature, t = 20 min 3Mass of the child, m = 30 kg = 30 × 10 gSpecific heat of the human body = Specific heat of water = c= 1000 cal/kg/ °C –1Latent heat of evaporation of water, L = 580 cal g The heat lost by the child is given as:∆θ = mc∆T= 30 × 1000 × (101 - 98) × (5/9)= 50000 calLet m1 be the mass of the water evaporated from the child’sbody in 20 min.Loss of heat through water is given by: ∆θ = m1L∴ m1 = ∆θ / L= (50000 / 580) = 86.2 g∴ Average rate of extra evaporation caused by the drug = m1 / t= 86.2 / 200= 4.3 g/min.11.19. A ‘thermacole’ icebox is a cheap and efficient method for storing smallquantities of cooked food in summer in particular. A cubical icebox of side 30 cmhas a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount ofice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermocole is 0.01 J –1 –1 –1 [Heat of fusion of s m K. 3 –1water = 335 × 10 J kg ]AnswerSide of the given cubical ice box, s = 30 cm = 0.3 m Thickness of the ice box, l = 5.0 cm = 0.05 mMass of ice kept in the ice box, m = 4 kg Time gap, t = 6 h = 6 × 60 × 60 s Outside temperature, T = 45°C –1Coefficient of thermal conductivity of thermacole, K = 0.01 J sPDFmyURL lets you convert a complete website to PDF automatically!
m–1 K–1 3 –1Heat of fusion of water, L = 335 × 10 J kg ’Let m be the total amount of ice that melts in 6 h. The amount of heat lost by the food:θ = KA(T - 0)t / l Where,A = Surface area of the box = 2 = 6 × 2 = 0.54 3 θ = 0.01 × 0.54 × 45 × 6 × 60 × 60 / 0.05 = 104976 6s (0.3) mJ But θ = m'L∴ m' = θ/L= 104976/(335 × 3 = 0.313 kg Mass of ice left = 4 – 0.313 = 3.687 kg 10 )Hence, the amount of ice remaining after 6 h is 3.687 kg. 211.20. A brass boiler has a base area of 0.15 m and thickness 1.0 cm. It boils water at the rate of 6.0 kg/minwhen placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler.Thermal conductivity of brass = 109 J s–1 –1 –1 3 –1 m K ; Heat of vaporisation of water = 2256 × 10 J kg .Answer 2Base area of the boiler, A = 0.15 m Thickness of the boiler, l = 1.0 cm = 0.01 m Boiling rate of water, R = 6.0kg/min Mass, m = 6 kgTime, t = 1 min = 60 s –1 –1 –1Thermal conductivity of brass, K = 109 J s–1 m K 3Heat of vaporisation, L = 2256 × 10 J kgThe amount of heat flowing into water through the brass base of the boiler is given by:θ = KA(T1 - T2) t / l ....(i)where,T1 = Temperature of the flame in contact with the boiler T2 = Boiling point of water = 100°CHeat required for boiling the water:θ = mL … (ii)Equating equations (i) and (ii), we get:∴ mL = KA(T1 - T2) t / l T1 - T2 = mLl / KAt 3= 6 × 2256 × 10 × 0.01 / (109 × 0.15 × 60) o= 137.98 CTherefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.11.21. Explain why:A.a body with large reflectivity is a poor emitterB.a brass tumbler feels much colder than a wooden tray on a chilly dayC.an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiationgives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value forthe temperature when the same piece is in the furnaceD.the earth without its atmosphere would be inhospitably coldE.heating systems based on circulation of steam are more efficient in warming a building thanthose based on circulation of hot water
AnswerA. A. a body with a large reflectivity is a poor absorber of light radiations. A poor absorber will in turn be apoor emitter of radiations. Hence, a body with a large reflectivity is a poor emitter.B.Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body tothe brass tumbler easily. Hence, the temperature of the body reduces to a lower value and one feels cooler.Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is conducted from thebody to the wooden tray. Hence, there is only a negligible drop in the temperature of the body and one doesnot feel cool.Thus, a brass tumbler feels colder than a wooden tray on a chilly day.C.An optical pyrometer calibrated for an ideal black body radiation gives too low a value for temperature ofa red hot iron piece kept in the open.Black body radiation equation is given by: (T4 4= = σ - T0 ) Where,= = Energy radiationT = Temperature of optical pyrometer To = Temperature of open spaceσ = ConstantHence, an increase in the temperature of open space reduces the radiation energy.When the same piece of iron is placed in a furnace, the radiation energy, E = σ 4 T(d) Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric gases, no extra heat willbe trapped. All the heat would be radiated back from earth’ssurface.(e) A heating system based on the circulation of steam is more efficient in warming a building than that based onthe circulation of hot water. This is because steam contains surplus heat in the form of latent heat (540 cal/g).Q. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from60 °C to 30 °C. The temperature of the surroundings is 20 °C.Ans.According to Newton’s law of cooling, we have: (-dT/T) = K(T - T0)dT / K(T - T0) = -Kdt ....(i) Where,Temperature of the body = TTemperature of the surroundings = T0 = 20°CK is a constantTemperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 sIntegrating equation (i), we get:
PDFmyURL lets you convert a complete website to PDF automatically!Q. A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C.If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat ofcombustion is 4.0 × 104 J/g?AnswerWater is flowing at a rate of 3.0 litre/min.The geyser heats the water, raising the temperature from 27°C to 77°C.Initial temperature, T1 = 27°C Final temperature, T2 = 77°C∴Rise in temperature, T = T2 – T1 = 77 – 27 = 50°CHeat of combustion = 4 × 104Specific heat of water, c = 4.2 JJ/gg–1 °C–1Mass of flowing water, m = 3.0 litre/min = 3000 g/min Total heat used, Q = mc T(e) 3000 × 4.2 × 50(f) 6.3 × 105 J/min∴ Rate of consumption = 6.3 × 105 / (4 × 104) = 15.75 g/min. Q. What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperatureby 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.)
Ans.Mass of nitrogen, m = 2.0 × 10–2 kg = 20 gRise in temperature, T = 45°CMolecular mass of N2, M = 28 J mol–1 K–1Universal gas constant, R = 8.3 Number of moles, n = m/M(h) (2 × 10-2 × 103) / 28(i) 0.714 Molar specific heat at constant pressure for nitrogen, Cp = (7/2)R= (7/2) × 8.3= 29.05 J mol-1 K-1 The total amount of heat to be supplied is given by the relation:Q = nCP T= 0.714 × 29.05 × 45= 933.38 JTherefore, the amount of heat to be supplied is 933.38 J.Q. Explain why(a) Two bodies at different temperatures T1 and T2 if brought inthermal contact do not necessarily settle to the mean temperature (T1 + T 2)/2.A.The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different partsof aB. plant from getting too hot) should have high specific heat.C.Air pressure in a car tyre increases during driving.D.The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.Ans.(a) In thermal contact, heat flows from the body at higher temprature to the body at lower temprature(b) till tempratures becomes equal. The final temprature can be the mean temprature (T1 + T2)/2only(c) when thermal capicities of the twobodies are equal.This is because heat absorbed by a substance is directly proportional to the specific heat of the substance.D.During driving, the temprature of air inside the tyre increases due to moion. Accordingto Charle's law, P ∝ T. Therefore, air pressure inside the tyre increases.This is because in a harbour town, the relative humidity is more than in a desert town. hence, theclimate of a harbour town is without extremes of hot and cold.Q. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas iscompressed to half its original volume?Ans.The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged betweensystem (cylinder) and its surroundings. Thus, the process is adiabatic.
Initial pressure inside the cylinder = P1 Final pressure inside the cylinder = P2 Initial volume inside the cylinder = V1 Final volume inside the cylinder = V2 Ratio of specific heats, γ = 1.4 For an adiabatic process, we have: P1V1γ = P2V2γ The final volume is compressed to half of its initial volume. ∴ V2 = V1/2 P1V1γ = P2(V1/2)γ P2/P1 = V1γ / (V1/2)γ = 2γ = 21.4 = 2.639 Hence, the pressure increases by a factor of 2.639. 12.5. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J) Answer The work done (W) on the system while the gas changes from state A to state B is 22.3 J. This is an adiabatic process. Hence, change in heat is zero. ∴Q = 0 W = –22.3 J (Since the work is done on the system) From the first law of thermodynamics, we have: Q= U+ W Where, U = Change in the internal energy of the gas ∴U = Q – W = – (– 22.3 J) U = + 22.3 J When the gas goes from state A to state B via a process, the net heat absorbed by the system is: Q = 9.35 cal = 9.35 × 4.19 = 39.1765 J Heat absorbed, Q = U + Q ∴ W= Q– U = 39.1765 – 22.3 = 16.8765 J Therefore, 16.88 J of work is done by the system. Q. Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:A. What is the final pressure of the gas in A and B?B. What is the change in internal energy of the gas?C. What is the change in the temperature of the gas?D. Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface? Answer Aa. When the stopcock is suddenly opened, the volume available to the gas at 1 atmospheric pressure will become two times. Therefore, pressure will decrease to one-half, i.e., 0.5 atmosphere. B.There will be no change in the internal energy of the gas as no work is done on/by the gas.
C. Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all. D.No, because the process called free expansion is rapid and cannot be controlled. the intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state. 12.7. A steam engine delivers 5.4×108 J of work per minute and services 3.6 × 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute? Ans. Work done by the steam engine per minute, W = 5.4 × 108 J Heat supplied from the boiler, H = 3.6 × 109 J Efficiency of the engine = Output energy / Input energy ∴ η = W / H = 5.4 × 108 / (3.6 × 109) = 0.15 Hence, the percentage efficiency of the engine is 15 %. Amount of heat wasted = 3.6 × 109 – 5.4 × 108 = 30.6 × 108 = 3.06 × 109 J Therefore, the amount of heat wasted per minute is 3.06 × 109 J. Q. An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing? Ans. Heat is supplied to the system at a rate of 100 W. ∴Heat supplied, Q = 100 J/s The system performs at a rate of 75 J/s. ∴Work done, W = 75 J/s From the first law of thermodynamics, we have: Q=U+W Where, U = Internal energy ∴U = Q – W= 100 – 75= 25 J/s= 25 W Therefore, the internal energy of the given electric heater increases at a rate of 25 W. 12.9. A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
Answer Total work done by the gas from D to E to F = Area of DEF Area of DEF = (1/2) DE × EF Where, DF = Change in pressure= 600 N/m2 – 300 N/m2= 300 N/m2 FE = Change in volume= 5.0 m3 – 2.0 m3= 3.0 m3 Area of DEF = (1/2) × 300 × 3 = 450 J Therefore, the total work done by the gas from D to E to F is 450 J.PDFmyURL converts any url to pdf Q. A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36° C, calculate the coefficient of performance. Ans. Temperature inside the refrigerator, T1 = 9°C = 282 K Room temperature, T2 = 36°C = 309 K Coefficient of performance = T1 / (T2 - T1) = 282 / (309 - 282) = 10.44 Therefore, the coefficient of performance of the given refrigerator is 10.44.13.KINETIC THEORYQ.Estimate the fraction of molecular volume to the actual volume occupiedby oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.Answer –8Diameter of an oxygen molecule, d = 3Å Radius, r = d/2 = 3/2 = 1.5 Å = 1.5 × 10 cm 3Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cmMolecular volume of oxygen gas, V = (4/3)πr3NWhere, N is Avogadro’s number = 6.023 × 1023 molecules/mole∴ V = (4/3) × 3.14 × (1.5 × -8 3 × 6.023 × 23 = 8.51 3 10 ) 10 cmRatio of the molecular volume to the actual volume of oxygen(g) 8.51 / 22400 -4(h) 3.8 × 10 .Q. Molar volume is the volume occupied by 1 mol of any (ideal) gas atstandard temperature and pressure (STP: 1 atmospheric pressure, 0 °C).Show that it is 22.4 litres.Ans.The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:PV = nRTWhere,
–1 –1 R is the universal gas constant = 8.314 J mol K n = Number of moles = 1 T = Standard temperature = 273 K 5 –2 P = Standard pressure = 1 atm = 1.013 × 10 Nm ∴ V = nRT / P 5 (j) 1 × 8.314 × 273 / (1.013 × 10 ) 3 (k) 0.0224 m (l) 22.4 litres Hence, the molar volume of a gas at STP is 22.4 litres. Q. Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures.A. What does the dotted plot signify?B.Which is true: T1 > T2 or T1 < T2?C.What is the value of PV/T where the curves meet on the y-axis?D.If we obtained similar plots for 1.00 ×10–3 kg of hydrogen, would we getthe same value of PV/T at the point where the curves meet on the y-axis?If not, what mass of hydrogen yields the same value of PV/T (for lowpressure high temperature region of the plot)? (Molecular mass of H2 =2.02μ, of O2 =32.0μ, R = 8.31 J mo1–1 K–1.) Answer =The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio PV/T is equal. μR (μ is the number of moles and R is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas. =The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature T1 is closer to the dotted plot than the curve of the gas at temperature T2. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, T1 > T2 is true for the given plot. (c) The value of the ratio PV/T, where the two curves meet, is μR. This is because the ideal gas equation is given as:
PV = μRTPV/T = μR Where,P is the pressureT is the temperature V is the volumeμ is the number of moles R is the universal constant –3 –1 –1Molecular mass of oxygen = 32.0 g Mass of oxygen = 1 × 10 kg = 1 g R = 8.314 J mole K -1∴ PV/T = (1/32) × 8.314 = 0.26 J KTherefore, the value of the ratio PV/T, where the curves meet on the y-axis, is –10.26 J K . –3(d) If we obtain similar plots for 1.00 × 10 kg of hydrogen, then we will not get the samevalue of PV/T at the point where the curves meet the y-axis. This is because the molecularmass of hydrogen (2.02 u) is different from that of oxygen (32.0 u). We have: -1PV/T = 0.26 J K –1 –1R = 8.314 J mole KMolecular mass (M) of H2 = 2.02 uPV/T = μR at constant temperature Where, μ = m/MConvert webpages to pdf online with PDFmyURL m = Mass of H2∴ m = (PV/T) × (M/R)= 0.26 × 2.02 / 8.31 –2 –5= 6.3 × 10 g = 6.3 × 10 kg –5Hence, 6.3 × 10 kg of H2 will yield the same value of PV/T.13.4. An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and atemperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressuredrops to 11 atm and its temperature drops to 17 °C. Estimate –1 –1the mass of oxygen taken out of the cylinder (R = 8.31 J mol K , molecular massof O2 = 32μ).Answer –3 3Volume of oxygen, V1 = 30 litres = 30 × 10 mGauge 5 pressure, P1 = 15 atm = 15 × 1.013 × 10 Pa Temperature, T1 = 27°C = 300 K –1 –1Universal gas constant, R = 8.314 J mole KLet the initial number of moles of oxygen gas in the cylinder be n1.The gas equation is given as:P1V1 = n1RT1∴ n1 = P1V1/ RT1
5 -3= (15.195 × 10 × 30 × 10 ) / (8.314 × 300) = 18.276 But n1 = m1 / MWhere,m1 = Initial mass of oxygenM = Molecular mass of oxygen = 32 g ∴m1 = n1M = 18.276 × 32 = 584.84 gAfter some oxygen is withdrawn from the cylinder, the pressure and temperature reduces. –3 3Volume, V2 = 30 litres = 30 × 10 m 5Gauge pressure, P2 = 11 atm = 11 × 1.013 × 10 Pa Temperature, T2 = 17°C = 290 KLet n2 be the number of moles of oxygen left in the cylinder. The gas equation is given as:P2V2 = n2RT2∴ n2 = P2V2/ RT2 5 -3= (11.143 × 10 × 30 × 10 ) / (8.314 × 290) = 13.86 But n2 = m2 / MWhere,m2 is the mass of oxygen remaining in the cylinder∴ m2 = n2M = 13.86 × 32 = 453.1 gThe mass of oxygen taken out of the cylinder is given by the relation:Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder= m1 – m2= 584.84 g – 453.1 g= 131.74 g= 0.131 kgTherefore, 0.131 kg of oxygen is taken out of the cylinder. 3Q. An air bubble of volume 1.0 cm rises from the bottom of a lake 40 m deep at atemperature of 12 °C. To what volume does it grow when it reaches the surface,which is at a temperature of 35 °C?Answer 3 –6 3Volume of the air bubble, V1 = 1.0 cm = 1.0 × 10 mBubble rises to height, d = 40 mTemperature at a depth of 40 m, T1 = 12°C = 285 K Temperature at the surface of the lake, T2= 35°C = 308 K The pressure on the surface of the lake: 5P2 = 1 atm = 1 ×1.013 × 10 PaThe pressure at the depth of 40 m: P1 = 1 atm + dρgWhere,ρ is the density of water = 3 3 10 kg/mg is the acceleration due to gravity = 9.8 2 ∴ P1 = 1.013 × 5 + 40 × 3 × 9.8 = 493300 Pa m/s 10 10We have P1V1 / T1 = P2V2 / T2Where, V2 is the volume of the air bubble when it reaches the surfaceV2 = P1V1T2 / T1P2 -6 5= 493300 × 1 × 10 × 308 / (285 × 1.013 × 10 )
–6 3 3= 5.263 × 10 m or 5.263 cm 3Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm .Q.Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapourand other constituents) in a 3room of capacity 25.0 m at a temperature of 27 °C and 1 atm pressure. 3Ans. Volume of the room, V = 25.0 m Temperature of the room, T = 27°C = 300 K 5Pressure in the room, P = 1 atm = 1 × 1.013 × 10 PaThe ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) canbe written as:PV = kBNTWhere, –23 2 –2 –1KB is Boltzmann constant = 1.38 × 10 m kg s K N is the number of air molecules in theroom∴ N = PV / kBT 5 -23= 1.013 × 10 × 25 / (1.38 × 10 × 300) 26= 6.11 × 10 molecules 26Therefore, the total number of air molecules in the given room is 6.11 × 10 .13.7. Estimate the average thermal energy of a helium atom at(i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) thetemperature of 10 million Kelvin (the typical core temperature in the case of a star).Answer(i) At room temperature, T = 27°C = 300 K Average thermal energy = (3/2)kT –23 2 –2 –1Where k is Boltzmann constant = 1.38 × 10 m kg s K ∴ (3/2)kT = (3/2) × 1.38 × -3810 × 300–21= 6.21 × 10 JHence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × –2110 J.(ii) On the surface of the sun, T = 6000 K Average thermal energy = (3/2) -38=(3/2) × 1.38 × 10 × 6000 -19= 1.241 × 10 JHence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × –1910 J. 7(iii) At temperature, T = 10 K Average thermal energy = (3/2)kT -23 7= (3/2) × 1.38 × 10 × 10 -16= 2.07 × 10 J
–16Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10 J.Q. Three vessels of equal capacity have gases at the same temperature and pressure.The first vessel contains neon (monatomic), the second contains chlorine (diatomic),and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equalnumber of respective molecules? Is the root mean square speed of molecules the samein the three cases? If not, in which case is vrms the largest?Ans.All the three vessels have the same capacity, they have the same volume.Hence, each gas has the same pressure, volume, and temperature.According to Avogadro’s law, the three vessels will contain an equal number of the respectivemolecules. This number isequal to Avogadro’s number, N = 6.023 × 23 10 .The root mean square speed (vrms) of a gas of mass m, and temperature T, is given by therelation: 1/2vrms = (3kT/m)where,k is Boltzmann constantFor the given gases, k and T are constants.Hence vrms depends only on the mass of the atoms, i.e., 1/2vrms ∝ (1/m)Therefore, the root mean square speed of the molecules in the three cases is not the same.Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest. Hence,neon has the largest root mean square speed among the given gases.Q.At what temperature is the root mean square speed of an atom in an argon gascylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar= 39.9 u, of He = 4.0 u).Ans.Temperature of the helium atom, THe = –20°C= 253 K Atomic mass of argon, MAr = 39.9u Atomic mass of helium, MHe = 4.0 u.Let, (vrms)Ar be the rms speed of argon. Let (vrms)He be the rms speed ofhelium. The rms speed of argon is given by:
3= 2523.675 = 2.52 × 10 K 3Therefore, the temperature of the argon atom is 2.52 × 10 K. Q.Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).Answer –7Mean free path = 1.11 × 10 m –1 9Collision frequency = 4.58 × 10 sSuccessive collision time ≈ 500 × (Collision time)Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 52.026 × 10 PaTemperature inside the cylinder, T = 17°C =290 K 10 10 10Radius of a nitrogen molecule, r = 1.0 Å = 1 × 10 m Diameter, d = 2 × 1 × 10 = 2 × 10 m –3Molecular mass of nitrogen, M = 28.0 g = 28 × 10 kg The root mean square speed ofnitrogen is given by the relation:Collision frequency = vrms / l -7 9 -1= 508.26 / (1.11 × 10 ) = 4.58 × 10 s Collision time is given as:T = d / vrms -10 -13= 2 × 10 / 508.26 = 3.93 × 10 s
Time taken between successive collisions: T ' = l / vrms -7 -10= 1.11 × 10 / 508.26 = 2.18 × 10 s -10 -13∴ T ' / T = 2.18 × 10 / (3.93 × 10 ) = 500Hence, the time taken between successive collisions is 500 times the time taken for a collision.Q. A metre long narrow bore held horizontally (and closed at one end) contains a 76 cmlong mercury thread, which traps a 15 cm column of air. What happens if the tube is heldvertically with the open end at the bottom? Ans. Length of the narrow bore, L = 1 m = 100 cm Length of the mercury thread, l = 76 cmLength of the air column between mercury and the closed end, la = 15 cmSince the bore is held vertically in air with the open end at the bottom, the mercury lengththat occupies the air space is: 100 – (76 + 15) = 9 cm.Hence, the total length of the aircolumn = 15 + 9 = 24 cm Let h cm of mercury flow out as a result of atmospheric pressure.∴Length of the air column in the bore = 24 + h cm And, length of the me3rcury column = 76 – h cmInitial pressure, P1 = 76 cm of mercury Initial volume, V1 = 15 cm 3Final pressure, P2 = 76 – (76 – h) = h cm of mercury Final volume, V2 = (24 + h) cmTemperature remains constant throughout the process.∴ P1V1 = P2V276 × 15 = h (24 + h) 2 + 24h – 1140 = 0 h= 23.8 cm or –47.8 cmHeight cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm ofmercury will remain in it. The length of the air column will be 24 + 23.8 = 47.8 cm.Q. From a certain apparatus, the diffusion rate of hydrogen 3 –1 has an average value of 28.7 cm s . The diffusion of another gas under the sameconditions is measured to have an average 3 –1rate of 7.2 cm s . Identify the gas.[Hint: Use Graham’s law of diffusion: R1/R2 = 1/2 where R1, R2 are diffusion rates of (M2/M1) ,gases 1 and 2, and M1 and M2 theirrespective molecular masses. The law is a simple consequence of kinetic theory.] 3 –1Ans. Rate of diffusion of hydrogen, R1 = 28.7 cm sRate of diffusion of another gas, R2 = 7.2 3 –1 According to Graham’s Law of diffusion, we have: cm s
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