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(B) Since the container is insulated,$Q = 0$. Since the partition is removed without doing any work,$W = 0$. According to the first law of thermodynam

Vedclass · Accepted Answer

$\frac{{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)}}{{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}$

Vedclass · Answer

(B) Since the container is insulated,$Q = 0$. Since the partition is removed without doing any work,$W = 0$. According to the first law of thermodynamics,$\Delta U = Q + W = 0$. This implies that the total internal energy remains constant: $U_{initial} = U_{final}$. For an ideal gas,the internal energy is $U = n C_v T$. Thus,$n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T$. Canceling $C_v$ from both sides,we get $T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$. Using the ideal gas law $PV = nRT$,we have $n_1 = \frac{P_1 V_1}{R T_1}$ and $n_2 = \frac{P_2 V_2}{R T_2}$. Substituting these into the expression for $T$: $T = \frac{(\frac{P_1 V_1}{R T_1}) T_1 + (\frac{P_2 V_2}{R T_2}) T_2}{\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}} = \frac{P_1 V_1 + P_2 V_2}{\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}}$. Simplifying the denominator: $T = \frac{R(P_1 V_1 + P_2 V_2)}{\frac{P_1 V_1 T_2 + P_2 V_2 T_1}{T_1 T_2}} = \frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1}$.

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