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Homework10-help - 16.24. Model: Treat the oxygen gas in the...

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16.24. Model: Treat the oxygen gas in the cylinder as an ideal gas. Solve: (a) The number of moles of oxygen is mol 50 g 1.563 mol 32 g mol M n M == = (b) The number of molecules is ( )( ) 23 1 23 A 1.563 mol 6.02 10 mol 9.41 10 Nn N × . (c) The volume of the cylinder Thus, () 2 22 0.10 m 0.40 m 1.257 10 m . Vr L ππ 3 23 25 3 23 9.41 10 7.49 10 m 1.257 10 m N V × × × (d) From the ideal-gas law pV = nRT we can calculate the absolute pressure to be ( )( ) 1.563 mol 8.31 J mol K 293 K 303 kPa 1.257 10 m nRT p V = × where we used T = 20 ° C = 293 K. But a pressure gauge reads gauge pressure : g 1 atm 303 kPa 101 kPa 202 kPa pp =− = = 16.19. Model: Treat the gas in the sealed container as an ideal gas. Solve: (a) From the ideal gas law equation pV = nRT , the volume V of the container is ( ) ( ) 3 5 2.0 mol 8.31 J mol K 273 30 K 0.0497 m 1.013 10 Pa nRT V p  +  = × Note that pressure must be in Pa in the ideal gas law. (b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is ( ) 12 2 21 1 273 130 K 1.0 atm 1.33 atm 273 pV T TT T + =⇒ = = = + Note that gas-law calculations must use T in kelvins 16.33. Model: Assume that the gas is an ideal gas.
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Homework10-help - 16.24. Model: Treat the oxygen gas in the...

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