JEE MAIN - Physics (2008 - No. 23)
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
$${{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)} \over {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}$$
$${{{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}} \over {{P_1}{V_1} + {P_2}{V_2}}}$$
$${{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}} \over {{P_1}{V_1} + {P_2}{V_2}}}$$
$${{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)} \over {{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}}}$$
Explanation
Since, no work is done and system is thermally insulated from surrounding. Therefore, total internal energy is constant, that is, $U=U_1+U_2$ Assuming gas have same degree of freedom, we have
$$ \left(n_1+n_2\right) C_{\mathrm{V}} T=n_1 C_{\mathrm{V}} T_1+n_2 C_{\mathrm{V}} T_2 $$
Therefore,
$$ T=\frac{n_1 R T_1+n_2 R T_2}{R n_1+n_2 R}=\frac{\left(P_1 V_1+P_2 V_2\right)}{\frac{P_1 V_1}{T_1}+\frac{P_2 V_2}{T_2}}=\frac{\left(P_1 V_1+P_2 V_2\right) T_1 T_2}{\left(P_1 V_1 T_2+P_2 V_2 T_1\right)} $$
$$ \left(n_1+n_2\right) C_{\mathrm{V}} T=n_1 C_{\mathrm{V}} T_1+n_2 C_{\mathrm{V}} T_2 $$
Therefore,
$$ T=\frac{n_1 R T_1+n_2 R T_2}{R n_1+n_2 R}=\frac{\left(P_1 V_1+P_2 V_2\right)}{\frac{P_1 V_1}{T_1}+\frac{P_2 V_2}{T_2}}=\frac{\left(P_1 V_1+P_2 V_2\right) T_1 T_2}{\left(P_1 V_1 T_2+P_2 V_2 T_1\right)} $$
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