Answer to Question #344445 in Inorganic Chemistry for Akhilesh chauhan

Question

Derive an expression for the entrophy change for isothermal mixing of ideal gas

Answer 1

The work done by the gas is W = ∫PdV.

PV = NkT. The temperature is constant so P = constant/V.

W = NkT∫(1/V)dV = NkT ln(V_f/V_i) = NkT ln(1 + V₂/V₁).

ΔU = ΔQ - ΔW. U = internal energy, Q = heat put into the system, W = work done by the system.

ΔU = 0, since the temperature is constant.

ΔQ = ΔW = NkT ln(1 + V₂/V₁).

ΔS = ΔQ_r/T = Nk ln(1 + V₂/V₁).