Two identical containers, each of volume V0, are joined by a small pipe of negligible volume. The co — States of Matter and Gaseous State Chemistry Question
Question
Two identical containers, each of volume V0, are joined by a small pipe of negligible volume. The containers contain identical gases at temperature T0 and pressure P0. One container is heated to temperature 2T0 while maintaining the other at same temperature, T0. The common pressure of the gas is P and n is the number of moles of gas in container at temperature 2T0, then
💡 Solution & Explanation
Total initial moles n_total = 2 × (P0 × V0 / R T0). After heating, the common pressure is P. Moles in the first container (at T0) n1 = P × V0 / R T0. Moles in the second container (at 2T0) n2 = P × V0 / R(2T0) = P × V0 / 2RT0. Since total moles are conserved: n1 + n2 = n_total → P V0 / R T0 × (1 + 1/2) = 2 P0 V0 / R T0 → 1.5 P = 2 P0 → P = 4/3 P0. Moles in the container at 2T0 are n = n2 = P V0 / 2RT0 = (4/3 P0) V0 / 2RT0 = 2P0V0/3RT0.