An insulated container of gas has two champers seperated 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

(a) [T₁T₂(P₁V₁+P₂V₂)] / (P₁V₁T₁+P₂V₂T₂)
(b) [T₁T₂(P₁V₁+P₂V₂)] / (P₁V₁T₂+P₂V₂T₁)
(c) (P₁V₁T₁+P₂V₂T₂) / (P₁V₁+P₂V₂)
(d) (P₁V₁T₂+P₂V₂T₁) / (P₁V₁+P₂V₂)

Correct Answer:

[T₁T₂(P₁V₁+P₂V₂)] / (P₁V₁T₂+P₂V₂T₁)

Explanation:

As this is a simple maxing of gas, even if adiabatic conditions are satisfied, PV = nRT for adiabatic as well as isothermal changes. The total number of molecules is conserved.
.·. n₁ = P₁V₁ / R₁T₁, n₂ = P₂V₂ / RT₂
Final state = (n₁ + n₂) RT
(n₁ + n₂) = (P₁V₁ / R₁T₁) + (P₂V₂ / RT₂) = [(T₂P₁V₁+T₁P₂V₂) / RT₁T₂]
T = [(T₁n₁ + T₂n₂) / (n₁ + n₂) = [T₁T₂(P₁V₁+P₂V₂)] / (P₁V₁T₂+P₂V₂T₁)