Derive Entropy of a mixture of Ideal Gases

For one mole of an ideal gas:
dS = C_v dT / T + R dV / V

Integrating the above equation, assuming that C_v remains constant for an ideal gas, we have:
S = C_v ln T + R ln V + S_o
Where S_o represents the integration constant

We know that C_p – C_v = R and V = R T / P, we will get:
S = C_p ln T – R ln R + R ln R + S_o
= C_p ln T – R ln P + S^’ _o
Where S^’ _o denotes the another constant

Entropy of a system which consists of a mixture of gases will be given by the sum of the individual entropies of the constituents at concentrations existing in the mixture. Let us suppose that n₁, n₂, n₃, etc., are the number of moles of the various gases which are present in the mixture. Let P₁, P₂, P₃, are their partial pressures respectively. Then the entropy of the mixture will be given by:
S = n₁ (C_p ln T – R ln P₁ + S^’ _o ) +
n₂ (C_p ln T – R ln P₂ + S^’ _o) +
n₃ (C_p ln T – R ln P₃ + S^’ _o)………..
= Σn (C_p ln T – R ln P + S^’ _o) …………………………………(1)

The partial pressure (p) of a n ideal gas is given by the following expression:
p = xP
Where x ——–> mole fraction of that particular gas in the mixture
P —————–> total pressure

Substituting the value of p = xP in equation (1), we will get:
S = Σn (C_p ln T – R ln P – R ln x + S^’ _o)

This is the entropy of a mixture of ideal gas