Question

Define entropy and show that it is a property of system.

Answer 1

Entropy is a thermodynamics property of a system which can be defined as the amount of heat contained in a substance and its interaction between two state in a process. Entropy increase with addition of heat and decrease when heat is removed. 

dQ = T.dS; T = Absolute Temperature and dS = Change in entropy. 

dS = dQ/ T 

T-S Diagrams

The area under T-S diagram represent the heat added or rejected. Entropy is a point function From first las

dQ = dU + dW

T.dS = dU + P.dV

Carnot efficiency η = (T₁– T₂)/T₁ = dW/dQ

dW = η.dQ; If T₁ – T₂ = 1; η =1/T

dW = dQ/T = dS; 

if Temperature difference is one. 

dS represents maximum amount of work obtainable per degree in temperature. 

Unit of Entropy = KJ/K 

Principle of Entropy 

From claucius inequality

∮ dQ T/ 0

Since dS = dQ/T for reversible process and dS > dQ/T for irreversible process

∮dQ T/ ≤ ∮ dS; or dQ/T d ≤ dS or dS e ≥ dQ/T

Change in Entropy During Process

1. V = C PROCESS

dQ = mCVdT

or, dQ/T = mCVdT/T

dS = mCVdT/T; or S₂ – S₁ = mCV ln P₂/P₁

2. P = C; PROCESS

dQ = mC_PdT

or, dQ/T = mC_PdT/T

dS = mC_PdT/T; or S₂ – S₁ = mCP lnT₂/T₁ = mCP ln V₂/V₁

3. T = C; PROCESS

dQ = mRT ln V₂/V₁

or, dQ/T = (mRT/T) ln V₂/V₁

or S₂ – S₁ = mR ln V₂/V₁ = m(CP – CV) ln V₂/V₁ = mR ln P₁/P₂

4. PV^y = C; PROCESS

dQ = 0; dS = 0

5. PVⁿ = C; PROCESS

dQ = [(γ – n)/ (γ – 1)] dW

= [(γ – n)/ (γ – 1)] PdV

dQ/T = [(γ – n)/ (γ – 1)] PdV/T

dS = [(γ – n)/ (γ – 1)] mRdV/V

or S₂ – S₁ = [(γ – n)/ (γ – 1)] mR ln V₂/V₁