Chapter 6, Problem 6.33P

(a)

Interpretation Introduction

Interpretation:

The heat absorbed by 1 kg of steam during a reversible isothermal expansion process from 800 kPa and $\text{200}\text{C}$ to 150 kPa

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of the system is given as:

$\begin{array}{l}\text{ΔU = q - w}\\ \text{where, q is the heat and w is the work done}\end{array}$

$\begin{array}{l}\text{Since this is an isothermal expansion process, }\Delta \text{U=0}\\ \text{q = w = nRTln}\frac{{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}}\\ {\text{here, V}}_{\text{1}}{\text{ and V}}_{\text{2}}\text{ are the initial and final volumes of steam}\\ \text{n = number of moles of steam}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The final temperature and work done by 1 kg of steam during a reversible adiabatic expansion process from 800 kPa and $\text{200}\text{C}$ to 150 kPa

Concept introduction:

Formula based of adiabatic reversible process

${\text{P}}_{\text{1}}{\text{V}}_{\text{1}}^{\text{γ}}{\text{ = P}}_{\text{2}}{\text{V}}_{\text{2}}^{\text{γ}}$

where,

$\text{γ = }\frac{\text{Cp}}{\text{Cv}}$

Work done during an adiabatic reversible process:

  $\text{W = }-[\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}{\text{-P}}_{\text{1}}{\text{V}}_{\text{1}}}{\text{γ-1}}\text{] }$