A bottle at 325K contains an ideal gas at a pressure of 162.5x10^3 Pa. The rubbe

Question

A bottle at 325K contains an ideal gas at a pressure of 162.5x10^3 Pa. The rubber stopper closing the bottle is removed. the gas expands adiabatically against Pexternal= 120.0x10^3 Pa, and some gas is expelled from the bottle in the process. When P=Pexternal the stopper is quickly replaced. The gas remaining in the bottle slowly warms up to 325K. What is the final pressure in the bottle for a monotomic gas, for which CV,m =3R/2, and a diatomic gas, for which Cv,m =5R/2?

Ideal gas. Work the problem twice: CV = 3R/2, then CV = 5R/2..
Initial T=325 K, P = 162.5 kPa, V0.
Adiabatic expansion against Pexternal = 120.0 kPa. Gas volume V0 is retained. Calculate the low temperature reached at the end of adiabatic expansion. Calculate the final pressure after constant-volume warming to 325 K.

Explanation / Answer

A bottle at 285K contains an ideal gas at a pressure of155.7*103Pa. The rubber stopper closing the bottle isremoved. The gas expands adiabatically against Pexternal=111.4*103Pa. and some gas isexpelled from the bottle in the process. whenP=Pexternal, The stopper is quickly replaced. the gasremaining in the bottle slowely warms up to 285K. what is the finalpressure in the bottle for a monatomic gas, for whichCv,m=3R/2, and a diatomic gas, whichCv,m=5R/2?

initial T = 285 K, pressure P = 155.7*103 Pa
after the adiabatic process, P' = 111.4*103 Pa, thetemperature = T'
P1-?T? = constant
P1-?T? =P'1-?T'?
?T/T' = (P'/P)(1-?)/?
After the stopper is replaced, for the gas inside the bottle,
initial pressure = P', temperature = T'
final pressure = P" = ?, temperature = T
it is a process with constant volume
P'/T' = P"/T
?P" = P'(T/T') = P'(P'/P)(1-?)/? =P'1/? P(?-1)/? =127.4*103 Pa

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