A bottle at 285 K contains an ideal gas at a pressure of 155.7 x10 3 Pa. The rub

Question

A bottle at 285 K contains an ideal gas at a pressure of 155.7 x103 Pa. The rubber stopper closing the bottleis removed. The gas expands adiabatically againstPexternal = 111.4 x 103 Pa.and some gas is expelled from the bottle in the process.
When P = Pexternal, the stopper is quicklyreplaced. The gas remaining in the bottle slowly warms up to285 K. What is the final pressure int he bottle for amonoatomic gas, for which Cv, m = 3R/2 and adiatomic gas for which  Cv, m  =5R/2 ?
Thanks!
When P = Pexternal, the stopper is quicklyreplaced. The gas remaining in the bottle slowly warms up to285 K. What is the final pressure int he bottle for amonoatomic gas, for which Cv, m = 3R/2 and adiatomic gas for which  Cv, m  =5R/2 ?
Thanks!

Explanation / Answer

I will make a guess! P1V1 =P2V2 for an ideal gasadiabatic expansion where = Cp/Cv =5/3 for monatomic So V2/V1 =(P1/P2)1/ The number of moles left in the bottle must beN2/N1 = V1/V2 The temperature of the gas left in the bottle can bedetermined from the ideal gas law T2/T1 =N1P2V2 /N2P1V1= (P2/P1)(V2/V1)2= (P1/P2)(2/)-1 Now that you have the temperature of the gas in thebottle after the expansion you can calculate the final temperature afterheating Then use = 7/5 for the diatomic gas Now that you have the temperature of the gas in thebottle after the expansion you can calculate the final temperature afterheating Then use = 7/5 for the diatomic gas

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