Answer to Question #344445 in Inorganic Chemistry for Akhilesh chauhan

Question #344445

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

1
Expert's answer
2022-05-25T06:28:01-0400

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(Vf/Vi) = 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 = ΔQr/T = Nk ln(1 + V2/V1). 


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