Answer: Option C
:
C
According to the kinetic theory, the average kinetic energy(KE) per molecule of a gas
= $\frac{3}{2}KT$. Let $n_1$ and $n_2$ be the number of moles of air in vessels 1 and 2 respectively.
Before mixing, the total KE of molecules in the two vessels is
$E_1=\frac{3}{2}{n}_{1}k{T}_1+\frac{3}{2}{n}_{2}k{T}_2$
= $\frac{3}{2}k(n_1{T}_1+n_2{T}_2)$
After mixing, the total KE of molecules is
$E_2=\frac{3}{2}(n_1+n_2)kT$
Where T is the temperature when equilibrium is established. Since there is no loss of energy
(Because the vessels are insulated), $E_2=E_1$ or
$\frac{3}{2}(n_1+n_2)kT=\frac{3}{2}k(n_1{T}_1+n_2{T}_2)$
or $T=\frac{n_1{T}_1+n_2{T}_2}{n_1+n_2}$
Now $P_1{V}_1=n_{1}R{T}_1$ and$P_2{V}_2=n_{2}R{T}_2$ which give
$n_1=\frac{P_1{V}_1}{R{T}_1}$ and $n_2=\frac{P_2{V}_2}{R{T}_2}$
Using these in Eq. (1) and simplifying, we get
$T=\frac{T_1{T}_2(P_1{V}_1+P_2{V}_2)}{(P_1{V}_1{T}_2+P_2{V}_2{T}_1)}$