1. Given two thermally insulated vessels containing the same ideal gas (air).

   Initial states:

   • Vessel 1: $(P_1, V_1, T_1)$
   • Vessel 2: $(P_2, V_2, T_2)$

   The vessels are rigid and insulated, so:

   • No heat exchange with surroundings: $Q=0$
   • Total volume remains constant: $V=V_1+V_2$
   • No external work is done

2. Use conservation of internal energy

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

   Since the system is isolated, $U_{\text{initial}}=U_{\text{final}}$

   Let final equilibrium temperature be $T$.

   Then, $n_1 C_V T_1 + n_2 C_V T_2 = (n_1+n_2) C_V T$

   Cancelling $C_V$, $n_1T_1+n_2T_2=(n_1+n_2)T$

   Hence, $T=\frac{n_1T_1+n_2T_2}{n_1+n_2}$

3. Express mole numbers using ideal gas equation

   $n_1=\frac{P_1V_1}{RT_1}, \qquad n_2=\frac{P_2V_2}{RT_2}$

   Substitute into the expression for $T$: $T=\frac{\left(\frac{P_1V_1}{RT_1}\right)T_1+\left(\frac{P_2V_2}{RT_2}\right)T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}$

4. Simplify

   In the numerator: $\left(\frac{P_1V_1}{RT_1}\right)T_1=\frac{P_1V_1}{R}, \qquad \left(\frac{P_2V_2}{RT_2}\right)T_2=\frac{P_2V_2}{R}$

   So, $T=\frac{\frac{P_1V_1+P_2V_2}{R}}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}$

   Cancelling $1/R$, $T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}$

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

5. Match with options

   This is exactly: $T_1T_2\frac{(P_1V_1+P_2V_2)}{P_1V_1T_2+P_2V_2T_1}$

   So the correct option is A.

6. Check other options briefly

   • B: $\frac{T_1+T_2}{2}$ only if mole numbers are equal, not in general.
   • C: dimensionally wrong as equilibrium temperature cannot be simple sum.
   • D: denominator is incorrect; it does not follow from energy conservation.

Final Answer: Option A.