Question

2. A bottle at 350 K contains an ideal gas at a pressure of 1.50x105 Pa. A rubber stopper is removed and you hear a pop as the gas expands quickly and adiabatically against Pext=1.00x105 Pa and some of the gas is expelled from the bottle. When P=Pext, the stopper is immediately replaced. The gas inside the bottle slowly equilibrates and goes back to having a temperature of 350K. What is the final pressure inside the bottle, assuming the gas is monoatomic (CV=3R/2)? What if the gas is diatomic (CV=5R/2)?

Answer 1

(P(V) ^G)= CONSTANT     FOR ADIABATIC PROCESS

G=5/3=1.66

In an adiabatic process, energy is transferred only as work. So, the gas is doing work on the environment if volume is expanding. The energy to do this work had to come from somewhere. It came from the internal energy of the gas. And hence the temperature of the gas goes down.

P1^(1-G)*(T1^G)=P2^(1-G)*(T2^G)

1.5^(-0.6)*350^1.6

SOLVING THIS WE GET TEMP OF 300.6 K TEMPERATURE.

AFTER WE HAVE TO GET TEMPERATURE OF 350 K TEMPERATURE.

P1/T1=P2/T2 FROM THIS

FINAL PRESSURE IS 1.16 *10^5