hw15_sol_s11

# hw15_sol_s11 - Phys 2101 Homework 15 Solution Spring `11...

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Phys 2101 Homework 15 Solution Spring `11 These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. 1. In solving the ideal-gas law equation pV = nRT for n , we first convert the temperature to the Kelvin scale: (40.0 273.15)K 313.15 K i T , and the volume to SI units: 3 3 3 1000 cm 10 m i V  . (a) The number of moles of oxygen present is       5 3 3 2 1.01 10 Pa 1.000 10 m 3.88 10 mol. 8.31J/mol K 313.15K i i pV n RT  (b) Similarly, the ideal gas law pV = nRT leads to        5 3 3 2 1.06 10 Pa 1.500 10 m 493K 3.88 10 mol 8.31J/mol K f f pV T nR  , which may be expressed in degrees Celsius as 220°C. Note that the final temperature can also be calculated by noting that ff ii if pV TT , or 53 1.06 10 Pa 1500 cm (313.15 K) 493 K 1.01 10 Pa 1000 cm fi    . 2. We assume that the pressure of the air in the bubble is essentially the same as the pressure in the surrounding water. If d is the depth of the lake and is the density of water, then the pressure at the bottom of the lake is p 1 = p 0 + gd , where p 0 is atmospheric pressure. Since p 1 V 1 = nRT 1 , the number of moles of gas in the bubble is n = p 1 V 1 / RT 1 = ( p 0 + gd ) V 1 / RT 1 , where V 1 is the volume of the bubble at the bottom of the lake and T 1 is the temperature there. At the surface of the lake the pressure is p 0 and the volume of the bubble is V 2 = nRT 2 / p 0 . We substitute for n to obtain       0 2 21 10 5 3 3 2 3 5 23 1.013 10 Pa + 0.998 10 kg/m 9.8m/s 40m 293K 20cm 277K 1.013 10 Pa 1.0 10 cm . p gd T VV Tp     3. 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 :

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hw15_sol_s11 - Phys 2101 Homework 15 Solution Spring `11...

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