A rigid insulated vessel is divided by a thin conducting partition. One side holds 2.00 mol of He at 300 K, the other 3.00 mol of N2 (rigid diatomic) at 460 K. After thermal equilibrium, the common temperature in K (one decimal) is

JEE Advanced levelNumerical answerDifficulty 3 of 3td034

A rigid insulated vessel is divided by a thin conducting partition. One side holds $2.00 mol$ of He at $300 K$ , the other $3.00 mol$ of $N X_{2}$ (rigid diatomic) at $460 K$ . After thermal equilibrium, the common temperature in K (one decimal) is

Numerical answer type, answer in K, tolerance 0.1.

Total energy is conserved with no work:

n_{1} C_{V 1} (T_{f} - 300) + n_{2} C_{V 2} (T_{f} - 460) = 0

with

C_{V} = \frac{3}{2} R

and

\frac{5}{2} R

T_{f} = (2 \times 1.5 \times 300 + 3 \times 2.5 \times 460) / (3 + 7.5) = 4350/10.5 = 414.29 K