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

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

$P_1, V_1, T_1 & P_2, V_2, T_2$

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container  therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

$-Q=W+U_1----1$

$Q=-W+U_2-----2$

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

$U_1 & U_2$ change in internal Energy of gas

Thus from 1 & 2 we can say that

$U_1+U_2=0$

$n_1c_v(T-T_1)+n_2c_v(T-T_2)=0$

$T(n_1+n_2)=n_1T_1+n_2T_2$

$T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}$

where $n_1=\frac{P_1V_1}{RT_1}$

$n_2=\frac{P_2V_2}{RT_2}$

$T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}$

$T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}$

and $P=\frac{P_1V_1+P_2V_2}{V_1+V_2}$