Two identical containers are connected by a thin tube with a valve. One container holds $1$ mole of helium at $300$ K, and the other holds $1$ mole of neon at $600$ K. When the valve is opened, the final equilibrium temperature is

Solution

1. Total internal energy $$\begin{array}{rl}U& =U_{He}+U_{Ne}\\ & =n_1{C}_{v1}{T}_1+n_2{C}_{v2}{T}_2\end{array}$$.
2. Since both are monatomic, $$\begin{array}{rl}{C}_{v1}& =C_{v2}\\ & =3R/2\end{array}$$.
3. Total energy $$\begin{array}{rl}U& =(3R/2)(1\times 300+1\times 600)\\ & =(3R/2)(900)\end{array}$$.
4. Final state $$\begin{array}{rl}{U}_f& =(n_1+n_2)C_v{T}_f\\ & =(2)(3R/2)T_f\\ & =3R{T}_f\end{array}$$.
5. Equating $3R{T}_f=(3R/2)(900)$, we get $T_f=450$ K.
6. Hence the answer is (B).