Q 11-12-004NEETNEET 2025Top questionMedium
A container has two chambers of volumes $V_1 = 2$ litres and $V_2 = 3$ litres separated by a partition made of a thermal insulator. The chambers contains $n_1 = 5$ and $n_2 = 4$ moles of ideal gas at pressures $p_1 = 1$ atm and $p_2 = 2$ atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of :
Answer: (B) $1.6$ atm
Concept. The isolated container exchanges no heat and does no work, so total internal energy $U = \dfrac{f}{2}pV$ is conserved (same gas on both sides).
Step 1. Energy conservation.
$$p(V_1 + V_2) = p_1V_1 + p_2V_2$$
Step 2. Solve.
$$p = \frac{1 \times 2 + 2 \times 3}{5} = 1.6\ \text{atm}$$
(The mole numbers are not needed.)
Answer. $1.6$ atm (Option B).
Solution by Sreeraj P, M.Sc Physics