Question

# A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal...

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.

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)

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