Question

A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and thermally conducting wall. Their pressure and volume are P1, V1 for part 1 and P2, V2 for part 2 respectively. Find the final pressure P and temperature T after the two gas reaches equilibrium. Assume the constant volume specific heats of the two gas are the same.

Answer 1

given rigid adiabatic container
two parts
n1, n2 moles of ideal gas
movable thermal conducting wall
P1, V1, P2, V2 for part 2
assuming constant Cv and same Cv for both

from ideal gas equation
P1V1 = n1RT1
P2V2 = n2RT2

final temperature = T
change in internal energy
U1 = n1Cv(T - T1) = n1*Cv*(T - P1V1/n1R)
U2 = n2*Cv*(T - P2V2/n2R)

also
final pressure will be same in both as well
hence
PV1' = n1*RT
PV2' = n2*RT

also
V1' + V2' = V1 + V2
hence
P = RT(n1 + n2)/(V1 + V2)
V1' = n1*(V1 + V2)/(n1 + n2)
V2' = n2*(V1 + V2)/(n1 + n2)
also
work done by gas 1 = W
work done by gas 2= -W
heat lost by gas 1 = Q
heat lost by gas 2 = -Q
hence

-Q = W + U1
Q = -W + U2

U1 + U2 = 0
hence
n1*Cv*(T - P1V1/n1R) + n2*Cv*(T - P2V2/n2R) = 0
n1*T - P1V1/R + n2*T - P2V2/R = 0
T = (P1V1 + P2V2)/R(n1 + n2)
and
P = RT(n1 + n2)/(V1 + V2) = (P1V1 + P2V2)/(V1 + V2)