General gas Equation for entropy change

Consider a certain quantity of perfect gas being heated by any thermodynamic process.

Let m = mass of gas

P₁ = initial pressure of gas

V₁ = initial volume of the gas

T₁ = initial temperature of the gas

And P₂ , V₂ ,T₂ are corresponding values for final condition.

The relation for change of entropy during the process may be expressed in following three ways.

1. In terms of volume and temperature

We know that for a state change from 1st law of thermodynamics

δQ = dU + W₁₋₂

dividing both sides with temperature

δQ = dU + W₁₋₂ -----(1)

T T T

now dU = mCvdT

W₁₋₂ = PdV

δQ = dS

T

dS = mCvdT + PdV -----(2)

T T

PV = mRT

P = mR

T V

Inserting this in equation (2)

dS = mCvdT + mRdV

T V

for whole process

ʃ dS = ʃ mCvdT + ʃ mRdV

T V

(S₂ - S₁) = mCv ln (T₂/T₁) + mR ln (V₂/V₁) -----(3)

2. In terms of pressure and temperature

We know that

(P₁V₁)/T₁ = (P₂V₂)/T₂

⇒V₂/V₁ = (P₁/P₂)×(T₂/T₁)

Inserting this value in (3)

(S₂ - S₁) = mCv ln (T₂/T₁) + mR ln [(P₁/P₂)×(T₂/T₁)]

(S₂ - S₁) = mCv ln (T₂/T₁) + mR ln (P₁/P₂)+ mR ln (T₂/T₁)

(S₂ - S₁) = m ln (T₂/T₁)×( Cv+R) + mR ln (P₁/P₂)

(S₂ - S₁) = mCp ln (T₂/T₁) + mR ln (P₁/P₂)

3. In terms of pressure and volume

We know that

(P₁V₁)/T₁ = (P₂V₂)/T₂

⇒T₂/T₁ = (P₂/P₁)×(V₂/V₁)

Insering this value in (3)

(S₂ - S₁) = mCv ln (P₂/P₁)×(V₂/V₁) + mR ln (V₂/V₁)

(S₂ - S₁) = mCv ln (P₂/P₁) + mCv ln (V₂/V₁) + mR ln (V₂/V₁)

(S₂ - S₁) = mCv ln (P₂/P₁) + m ln (V₂/V₁)×(Cv+R)

(S₂ - S₁) = mCv ln (P₂/P₁) + mCp ln (V₂/V₁)