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)

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