JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Heat & Thermodynamics

The vessel is closed and insulated, so the total internal energy of the two portions of the same monatomic gas is simply shared until they reach a common temperature, with no heat leaving the system. For a monatomic ideal gas $U=n{C}_{V}T$ with $C_V=\frac{3}{2}R$, and since the same $C_V$ appears everywhere the balance reduces to $n_1{T}_1+n_2{T}_2=(n_1+n_2)T_f$. Substituting the data, $(0.1)(200)+(0.05)(400)=(0.15)T_f$, that is $20+20=0.15T_f$, so $T_f=\frac{40}{0.15}\approx 266.67$ K.