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An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $V_1$ and contains ideal gas at pressure $P_1$ and temperature $T_1$. The other chamber has volume $V_2$ and contains ideal gas at pressure $P_2$ and temperature $T_2$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be
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Verified Answer
The correct answer is:
$\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_2+P_2 V_2 T_1}$
$\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_2+P_2 V_2 T_1}$
$$
\begin{aligned}
& U=U_1+U_2 \\
& T=\frac{\left(P_1 V_1+P_2 V_2\right) T_1 T_2}{\left(P_1 V_1 T_2+P_2 V_2 T_1\right)}
\end{aligned}
$$
\begin{aligned}
& U=U_1+U_2 \\
& T=\frac{\left(P_1 V_1+P_2 V_2\right) T_1 T_2}{\left(P_1 V_1 T_2+P_2 V_2 T_1\right)}
\end{aligned}
$$
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