An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V₁ and contains ideal gas at pressure P₁ and temperature T₁. The other chamber has volume V₂ and contains ideal gas at pressure P₂ and temperature T₂. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be (a) (T₁T₂(P₁V₁+P₂V₂))/(P₁V₁T₂+P₂V₂T₁) (b) (P₁V₁T₁+P₂V₂T₂)/(P₁V₁+P₂V₂) (c) (P₁V₁T₂+P₂V₂T₁)/(P₁V₁+P₂V₂) (d) (T₁T₂(P₁V₁+P₂V₂))/(P₁V₁T₁+P₂V₂T₂)

Answer: (a). Insulated and no work, so total internal energy is conserved: n₁CvT₁ + n₂CvT₂ = (n₁ + n₂)CvT. Moles: n₁ = (P₁V₁)/(RT₁), n₂ = (P₂V₂)/(RT₂). T = (P₁V₁ + P₂V₂)/((P₁V₁)/(T₁) + (P₂V₂)/(T₂)) = (T₁T₂(P₁V₁ + P₂V₂))/(P₁V₁T₂ + P₂V₂T₁) So the answer is (a).

An insulated container of gas has two chambers separated by an… | JEE Main 2008 (AIEEE)

Q80JEE Main 2008 · AIEEEDifficult

Gas Laws & Ideal Gas Equation

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $V_1$ and contains ideal gas at pressure $P_1$ and temperature $T_1$ . The other chamber has volume $V_2$ and contains ideal gas at pressure $P_2$ and temperature $T_2$ . If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

Show answer and solution

Answer: (a)

Insulated and no work, so total internal energy is conserved: $n_1C_vT_1 + n_2C_vT_2 = (n_1 + n_2)C_vT$ .
Moles: $n_1 = \frac{P_1V_1}{RT_1}$ , $n_2 = \frac{P_2V_2}{RT_2}$ .
$T = \frac{P_1V_1 + P_2V_2}{\frac{P_1V_1}{T_1} + \frac{P_2V_2}{T_2}} = \frac{T_1T_2(P_1V_1 + P_2V_2)}{P_1V_1T_2 + P_2V_2T_1}$
So the answer is (a).

Report a mistake

Gas Laws & Ideal Gas Equation question from JEE Main 2008 (AIEEE)

More questions on Gas Laws & Ideal Gas Equation