ssm_ch19 - Chapter 19 Student Solutions Manual 7. (a) In...

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Chapter 19 – Student Solutions Manual 7. (a) In solving pV = nRT for n , we first convert the temperature to the Kelvin scale: . And we convert the volume to SI units: 1000 cm (40.0 273.15)K 313.15 K T =+ = 3 = 1000 × 10 –6 m 3 . Now, according to the ideal gas law, () ( ) ( ) 56 3 2 1.01 10 Pa 1000 10 m 3.88 10 mol. 8.31J/mol K 313.15K pV n RT ×× == (b) The ideal gas law pV = nRT leads to ( ) 3 2 1.06 10 Pa 1500 10 m 493K. 3.88 10 mol 8.31J/mol K pV T nR = ×⋅ We note that the final temperature may be expressed in degrees Celsius as 220°C. 13. Suppose the gas expands from volume V i to volume V f during the isothermal portion of the process. The work it does is ln , ff ii VV f i V dV W p dV nRT nRT = ∫∫ where the ideal gas law pV = nRT was used to replace p with nRT / V . Now V i = nRT / p i and V f = nRT / p f , so V f / V i = p i / p f . Also replace nRT with p i V i to obtain ln . i f p Wp V p = Since the initial gauge pressure is 1.03 × 10 5 Pa, p i = 1.03 × 10 5 Pa + 1.013 × 10 5 Pa = 2.04 × 10 5 Pa. The final pressure is atmospheric pressure: p f = 1.013 × 10 5 Pa. Thus ( ) 5 53 5 2.04 10 Pa 2.04 10 Pa 0.14m ln 2.00 10 J. 1.013 10 Pa W ⎛⎞ × = × ⎜⎟ × ⎝⎠ 4 During the constant pressure portion of the process the work done by the gas is W = p f ( V i V f ). The gas starts in a state with pressure p f , so this is the pressure throughout this portion of the process. We also note that the volume decreases from V f to V i . Now V f = p i V i / p f , so
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() 55 4 1.013 10 Pa 2.04 10 Pa 0.14m 1.44 10 J. ii fi f i i f pV Wp V pp V p ⎛⎞ =− = = ×− × ⎝⎠ × ( ) 3 The total work done by the gas over the entire process is W = 2.00 × 10 4 J – 1.44 × 10 4 J = 5.60 × 10 3 J. 19. According to kinetic theory, the rms speed is rms 3 RT v M = where T is the temperature and M is the molar mass. See Eq. 19-34. According to Table 19-1, the molar mass of molecular hydrogen is 2.02 g/mol = 2.02 × 10 –3 kg/mol, so ( ) 2 rms 3 3 8.31J/mol K 2.7K 1.8 10 m/s. 2.02 10 kg/mol v == × × 21. Table 19-1 gives M = 28.0 g/mol for nitrogen. This value can be used in Eq. 19-22 with T in Kelvins to obtain the results. A variation on this approach is to set up ratios, using the fact that Table 19-1 also gives the rms speed for nitrogen gas at 300 K (the value is 517 m/s). Here we illustrate the latter approach, using v for v rms : 2 22 11 1 3/ . RT M vT RT M (a) With T 2 = (20.0 + 273.15) K 293 K, we obtain 2 293K 517m/s 511m/s. 300K v (b) In this case, we set 1 3 2 v = 2 v and solve 32 / vv TT = / for T 3 : 2 2 3 2 1 73.0K 2 v v = ⎜⎟ which we write as 73.0 – 273 = – 200°C.
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ssm_ch19 - Chapter 19 Student Solutions Manual 7. (a) In...

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