Chapter 5, Problem 5.12P

Interpretation Introduction

Interpretation:

Prove that for an ideal gas,

  $\frac{\Delta S}{R}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_V{}^{ig}}{R}\frac{dT}{T}+\ln \frac{V}{V_0}}$

Concept Introduction:

The entropy change of process is defined as:

  $dS=\frac{d{Q}_{reversible}}{T}$

And from first law of thermodynamics,

  $d{Q}_{reversible}=dU+PdV$......(1)

For ideal gas

  $\begin{array}{l}PV=RT\\ \text{and}\\ dH=C_p{}^{ig}dT\\ \text{and}\\ C_P-R=C_V\end{array}$