Please see the attached file for full description of the problem concerning volume and pressure of gases in a thermally isolated container.© BrainMass Inc. brainmass.com March 4, 2021, 5:51 pm ad1c9bdddf
For monoatomic gas: r (gamma) = 5/3 = 1.67
and, 1/r = 3/5 = 0.6
Because, no heat energy is transferred, hence, these processes will be adiabatic processes.
dQ = dU + dW
dQ = 0
dU = -dW
For gas A:
PA*VA/TA = PA'*VA'/TA' .....(1)
PA*VA^r = PA'*VA'^r ....(2)
where, PA, VA and TA are the pressure, volume and temperature of gas A before unlock and PA', VA', and TA' are at the equilibrium.
Similarlly for gas B:
PB*VB/TB = PB'*VB'/TB' .....(3)
PB*VB^r = PB'*VB'^r ....(4)
At equilibrium: PA' = PB' ....(5)
The solution explains clearly how to find various things about the volume and gas and how they and their equilibria changes in a thermally isolated container according to how the partition in the middle of it moves.