Question

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 ______.

Options

• `(T_1T_2(P_1V_1 + P_2V_2))/(P_1V_1T_2 + P_2V_2T_1)`

• `(P_1V_1T_1 + P_2V_2T_2)/(P_1V_1 + P_2V_2)`

• `(P_1V_1T_2 + P_2V_2T_1)/(P_1V_1 + P_2V_2)`

• `(T_1T_2(P_1V_1 + P_2V_2))/(P_1V_1T_1 + P_2V_2T_2)`

Answer 1

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 `underlinebb((T_1T_2(P_1V_1 + P_2V_2))/(P_1V_1T_2 + P_2V_2T_1))`.

Explanation:

Here Q = 0 and W = 0. Therefore from first law of thermodynamics ΔU = Q + W = 0.

Internal energy of first vessle + Internal energy of second vessel = Internal energy of combined vessel

`n_1C_νT_1 + n_2C_νT_2 = (n_1 + n_2)C_νT`

⇒ T = `(n_1T_1 + n_2T_2)/(n_1 + n_2)`

For first vessel `n_1 = (P_1V_1)/(RT_1)`, for second vessel `n_2 = (P_2V_2)/(RT_2)`

∴ T = `((P_1V_1)/(RT_1) xx T_1 + (P_2V_2)/(RT_2) xx T_2)/((P_1V_1)/(RT_1) + (P_2V_2)/(RT_2))`

= `(T_1T_2(P_1V_1 + P_2V_2))/(P_1V_1T_2 + P_2V_2T_1)`