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Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases Additional end of chapter problems E1.2(c) A perfect gas undergoes isothermal compression, which reduces its volume by 3.08 dm3. The final pressure and volume of the gas are 6.42 bar and 5.38 dm3, respectively. Calculate the original pressure of the gas in (a) bar, (b) atm. E1.2(c) Boyle’s law [1.6] in the form pf Vf piVi can be solved for either initial or final pressure, hence pi Vf pf Vi Vf 5.38dm3 Vi 5.38dm3 3.08dm 3 8.46 dm 3 p f 6.42bar Therefore, 5.38dm3 (a) pi 3 (6.42bar) 4.08bar 8.46dm 1atm 4.03atm (b) Since 1 atm = 1.013 bar, pi (4.08bar) 1.013bar E1.3(c) A car tyre (i.e. an automobile tire) was inflated to a pressure of 30 lb in –2 (1.00 atm = 14.7 lb in–2) on a winter’s day when the temperature was 0F. What pressure will be found, assuming no leaks have occurred and that the volume is constant, on a subsequent hot summer’s day when the temperature is 95F? What complications should be taken into account in practice? E1.3(c) The perfect gas law, pV nRT [1.8], can be rearranged to constant. Hence, p nR constant , if n and V are T V pf pi T or, solving for pf , pf f pi Tf Ti Ti Internal pressure pump pressure atmospheric pressure 2 2 2 pi 30 lb in 147 lb in 447 lb in Ti 255 K(0F) Tf 308 K(95F) pf 308K 2 2 447 lbin 540 lbin 255K 2 2 Therefore, p(pump) 540 lbin 147 lbin 39 lbin 2 Complications are those factors that destroy the constancy of V or n, such as the change in volume of the tyre, the change in rigidity of the material from which it is made, and loss of pressure by leaks and diffusion. E1.5(c) What pressure difference must be generated along a vertical single stage water pump tube to extract water from a well 5.0 m deep? E1.5(c) Identifying pex in the equation p pex gh [13] as the pressure at the top of the tube and p as the atmospheric pressure on the liquid, the pressure difference is OXFORD © Oxford University Press, 2006. All rights reserved. Higher Education Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases Additional end of chapter problems 3 p pex gh (10 103 kg m ) (981m s 2) (5.0 m) 5.0 104 Pa ( 0.50 atm) E1.6(c) A manometer consists of a U-shaped tube containing a liquid. One side is connected to the apparatus and the other is open to the atmosphere. The pressure inside the apparatus is then determined from the difference in heights of the liquid. Suppose the liquid is an oil with density 0.865 g cm-3, the external pressure is 750 Torr, and the open side is 15.0 cm lower than the side connected to the apparatus. What is the pressure in the apparatus? E1.6(c) The pressure in the apparatus is given by p patm gh[1.3] 1 atm 1.013105 Pa 0.9997 10 5 Pa p 750Torr atm 1 atm 760Torr gh 0.865 g cm -3 1 kg 3 10 g 106 cm3 3 m -2 9.806 m s 0.15 m 1272 Pa 5 3 5 p 0.9997 10 Pa + 1.272 10 Pa = 1.012 10 Pa = 101 kPa E1.8(c) At 0.000C and 1.000 atm, the mass density of the vapour of a hydrocarbon is 1.97 kg m–3. What is the molecular formula of the hydrocarbon under these conditions? What is the hydrocarbon? E1.8(c) Since p 1 atm , the approximation that the vapour is a perfect gas is adequate. Then (as in Exercise 1.7(b)), m pV nRT RT Upon rearrangement M (8.314 Pa m3 K 1 mol1 ) (273K) RT (1.97 kg m3 ) 0.0441k g mol 1= 44.1 g mol-1 5 p 1.013 10 Pa M The formula of the vapour is then C3 H 8 , which is one of the propanes. E1.9(c) Calculate the mass of water vapour present in a house of volume 1000 m3 that contains air at 22C on a day when the relative humidity is 48 per cent indoors. OXFORD © Oxford University Press, 2006. All rights reserved. Higher Education Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases Additional end of chapter problems E1.9(c) The partial pressure of the water vapour in the house is: pH2O (048) (`19.827Torr) 9.5Torr Assuming that the perfect gas equation [1.8] applies, with n m pV m RT or M M 3 3 1atm 10 dm 1 (1000 m3 ) (9.5Torr) m3 (1802 g mol ) pVM 760 Torr m RT 1 (00821 dm3 atm K mol 1) (295K) 9.3 10 g 9.3 kg 3 E1.12(c) For dry air at 1.0000 atm pressure, the densities at –50C, 0C, and 69C are 1.5826 g dm–3, 1.2929 g dm–3, and 1.0322 g dm–3, respectively. From these data, and assuming that air obeys Charles’s law, determine a value for the absolute zero of temperature in degrees Celsius. E1.12(c) The easiest way to solve this exercise is to assume a sample of mass 1000g , then calculate the volume at each temperature, plot the volume against the Celsius temperature, and extrapolate to V 0. Draw up the following table / C /(g dm -3 ) V/(dm3 g -1 ) -50 1.5826 0.6319 0 1.2929 0.7734 69 1.0322 0.9688 Linear regression of the V versus data gives a value for absolute zero close to 273 C . Alternatively, one could use an equation for V as a linear function of , which is Charles’s law, and solve for the value of absolute zero. V V0 (1 ) At absolute zero, V 0 , then (abszero) one of the data points (except 0 ) as follows. From V V0 (1 ) , 1 . The value of can be obtained from any V 09688 V 1 07734 1 0 0003661(C) 1 100C 1 1 273C 0003661(C) 1 which is the same to within experimental error as the value obtained graphically. E1.13(c) Calculate the pressure exerted by 1.0 mol C6H6 behaving as (a) a perfect gas, (b) a van der Waals gas when it is confined under the following conditions: (i) at 373.15 K in 22.414 dm3, (ii) at 1000 K in 300 cm3. Use the data in Table 1.6. E1.13(c) (a) p nRT [18] V OXFORD © Oxford University Press, 2006. All rights reserved. Higher Education Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases Additional end of chapter problems n 10 mol T 37315K (i) or 1000 K (ii) V 22414 dm (i) or 300 cm 3 (ii) 3 1 p (i) (10 mol) (8206 102 dm3 atm K mol 1) (37315K) 1.37 atm 22414dm3 1 (ii) p (10 mol) (8206 102 dm3 atm K mol 1) (1000 K) 2.73 102 atm 0300dm3 2 p nRT an2 [121a ] V nb V (b) 1 From Table 1.5, a 5507dm6 atmmol2 and b 651 102 dm3 mol . Therefore, 1 2 1 3 nRT (10 mol) (8206 10 dm atm K mol ) (37315 K) 137 atm V nb [22414 (10) (651102 )]dm3 (i) 2 an 2 (5507 dm6 atm mol ) (10 mol) 2 111102 atm V2 (22414 dm3 ) 2 and p 137 atm 111102 atm 1.36atm 1 (ii) 2 1 3 nRT (10 mol) (8206 10 dm atm K mol ) (1000 K) V nb (0300 00651) dm 3 349 102 atm 2 an 2 (5507 dm6 atm mol ) (10 mol) 2 1.83 102 atm V2 (0300 dm3 ) 2 and p 3.49 102 atm 1.84 102 atm 1.65 102 atm Comment. It is instructive to calculate the percentage deviation from perfect gas behaviour for (i) and (ii). (i) 1.36 1.37 100% 0.7% 1.37 (ii) (1.65 102 ) (2.73 102 ) 100% 39.5% 2.73 102 Deviations from perfect gas behaviour are not observed at p 1atm except with very precise apparatus. E1.16(c) Cylinders of compressed gas are typically filled to a pressure of 200 bar. For nitrogen, what would be the molar volume at this pressure and 25C based on (a) the perfect gas equation, (b) the van der Waals equation. For nitrogen, a = 1.39 dm6 atm mol–2, b = 0.0391 dm3 mol–1. E1.16(c) (a) Vmo 1 RT (8314 J K mol 1) (29815 K) 1 p (200 bar) (105 Pa bar ) 124 104 m 3 mol 1 0.124dm3 mol 1 OXFORD © Oxford University Press, 2006. All rights reserved. Higher Education Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases Additional end of chapter problems (b) The van der Waals equation is a cubic equation in Vm . The most direct way of obtaining the molar volume would be to solve the cubic analytically. However, this approach is cumbersome, so we proceed as in Example 1.6. The van der Waals equation is rearranged to the cubic form RT 2 a ab RT 2 a ab Vm3 b 0 or x3 b 0 Vm Vm x x p p p p p p 1 with x Vm (dm3 mol ) . The coefficients in the equation are evaluated as b+ (8.206´ 10- 2 dm3 mol- 1 )´ (298.15K ) RT = (3.91´ 10- 2 dm3 mol- 1 ) + p (200 bar )´ (1.013 atm bar - 1 ) = (3.91´ 10- 2 + 0.1208)dm3 mol- 1 = 0.1599 dm 3mol- 1 a 1.39 dm6 atm mol- 2 = = 6.86´ 10- 3 (dm3 mol- 1 ) 2 - 1 p (200bar) ´ (1.013 atm bar ) ab (1.39 dm 6 atm mol- 2 ) ´ (3.91´ 10- 2 dm3 mol- 1 ) = = 2.68´ 10- 4 (dm3 mol- 1 )3 p (200 bar)´ (1.013 atm bar- 1 ) Thus, the equation to be solved is x3 - 0.1599x2 + (6.86´ 10- 3 ) x - (2.68´ 10- 4 ) = 0. Calculators and computer software for the solution of polynomials are readily available. In this case we find x 0122 or Vm 0.122 dm3 mol 1 Thedifferenceis about 2 percent. E1.18(c) A vessel of volume 24.4 dm3 contains 1.0 mol H2 and 2.5 mol N2 at 298.15 K. Calculate (a) the mole fractions of each component, (b) their partial pressures, and (c) their total pressure. n n(H 2 ) n(N 2 ) 10 mol 2.5 mol 35 mol E1.18(c) (a) (b) xJ nJ [114] n 10 mol 2.5mol 029 x(N 2 ) 071 35mol 35mol The perfect gas law is assumed to hold for each component individually as well as for the RT mixture as a whole. Hence, pJ nJ V x(H 2 ) RT V 1 (8206 102 dm3 atm K mol 1) (29815K) 1 100 atm mol 3 24.4dm 1 p(H2 ) (10 mol) (100 atm mol ) 10atm 1 p(N2 ) (2.5mol) (100 atm mol ) 2.5atm (c) p p(H 2 ) p(N 2 )[115] 10 atm 2.5atm 35atm E1.19 (c)The critical constants of carbon dioxide are pc = 72.85 atm, Vc = 94.0 cm3 mol–1, and Tc = 304.2 K. Calculate the van der Waals parameters of the gas and estimate the OXFORD © Oxford University Press, 2006. All rights reserved. Higher Education Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases Additional end of chapter problems radius of the molecules. Compare to the values in Table 1.6. What might your comparison suggest? E1.19(c) Equations [1.22] are solved for b and a, respectively, and yield b Vc and a 27b 2 pc 3Vc2 pc 3 Substituting the critical constants 1 b (94.0 cm3 mol 1) 31.3 cm3 mol 1 3 1 a 3 (94.0 103 dm3 mol ) 2 (72.85atm) 1.93dm6 atm mol 2 Note that knowledge of the critical temperature, Tc , is not required. As b is approximately the volume occupied per mole of particles 31.3 106 m3 mol 1 b vmol 520 1029 m3 NA 6022 1023 mol 1 1 3 4 3 (520 1029 m3 ) 023nm Then, with vmol r 3 r 3 4 The values of a and b given in Table 1.6 differ substantially from the values calculated here. Consequently, the value of the radius of the molecule given above cannot be taken too seriously. E1.20(c) Use the van der Waals parameters for carbon dioxide to calculate approximate values of (a) the Boyle temperature of the gas and (b) the radius of a CO2 molecule regarded as a sphere (a = 3.610 dm6 atm mol-2, b = 0.0429 dm3 mol-1). E1.20(c) (a) The Boyle temperature is the temperature at which lim Vm van der Waals equation Z so pVm RT RT Vm b Va2 Vm m RT dZ vanishes. According to the d(1Vm ) Vm a Vm b Vm RT dZ dZ d(1Vm ) dVm dVm d(1Vm ) dZ Vm 1 a 2 Vm2 2 Vm 2 dVm (Vm b) Vm b Vm RT Vm2 b a (Vm b) 2 RT In the limit of large molar volume, we have lim Vm dZ a b 0 so d(1Vm ) RT a b RT 2 and T (3.610dm6 atm mol ) a 1025K Rb (008206 dm3 atm K 1 mol 1) (00429 dm3 mol 1) (b) By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e. twice their radius); the Avogadro constant times the volume is the molar excluded volume b. OXFORD © Oxford University Press, 2006. All rights reserved. Higher Education Atkins & de Paula, Physical Chemistry 8e, Chapter 1: The properties of gases Additional end of chapter problems 1 3 4 (2r )3 1 3b b NA so r 3 2 4 N A 1 3 1 3(00429 dm 3 mol 1) r 2 4 (6022 1023 mol 1) 1286 109 dm 129 1010 m 0129 nm E1.22(c) A certain gas obeys the van der Waals equation with a = 0.580 m6 Pa mol–2. Its volume is found to be 3.50 10–4 m3 mol–1 at 273 K and 5.0 MPa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure? E1.22(c) The van der Waals equation [121b] is solved for b, which yields b Vm RT a p 2 Vm Substituting the data 1 b 3.50 104 m3 mol 1 (8314 J K mol 1) (273K) 0580 m 6 Pa mol 2 6 (5 0 10 Pa) 4 3 1 2 (3.50 10 m mol ) 1.17 104 m3 mol 1 Z pVm (50 106 Pa) (3.50 104 m3 mol-1 ) [1.17] 077 1 RT (8314 J K mol 1) (273K) Comment. The definition of Z involves the actual pressure, volume, and temperature and does not depend upon the equation of state used to relate these variables. OXFORD © Oxford University Press, 2006. All rights reserved. Higher Education