Key idea: insulated container and no work done

The container is insulated, so $Q=0$ and the partition is removed without doing external work, so $W=0.$

Therefore, for the total gas system, $\Delta U=Q-W=0.$

Internal energy of an ideal gas

For an ideal gas, internal energy depends only on temperature: $U=nC_VT.$

Hence, initial total internal energy equals final total internal energy.

Initial moles in the two chambers

Using the ideal gas equation, $n_1=\frac{P_1V_1}{RT_1}, \qquad n_2=\frac{P_2V_2}{RT_2}.$

Initial total internal energy

Assuming the same ideal gas is present in both chambers, with the same constant $C_V$, $U_i=n_1C_VT_1+n_2C_VT_2.$

Substitute $n_1,n_2$: $U_i=\frac{P_1V_1}{RT_1}C_VT_1+\frac{P_2V_2}{RT_2}C_VT_2$ $U_i=\frac{C_V}{R}(P_1V_1+P_2V_2).$

Final internal energy

After removing the partition, total moles are $n=n_1+n_2=\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}.$

If final equilibrium temperature is $T_f$, then $U_f=(n_1+n_2)C_VT_f.$

So, $U_f=C_VT_f\left(\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}\right).$

Apply conservation of internal energy

Since $U_i=U_f$, $\frac{C_V}{R}(P_1V_1+P_2V_2)=C_VT_f\left(\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}\right).$

Cancel $\frac{C_V}{R}$: $P_1V_1+P_2V_2=T_f\left(\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}\right).$

Therefore, $T_f=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}.$

Multiply numerator and denominator by $T_1T_2$: $T_f=\frac{T_1T_2(P_1V_1+P_2V_2)}{P_1V_1T_2+P_2V_2T_1}.$

Match with options

This matches Option A: $\boxed{\frac{T_1T_2\left(P_1V_1+P_2V_2\right)}{P_1V_1T_2+P_2V_2T_1}}$

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.