Chapter 5, Problem 5.15NP

Interpretation Introduction

Interpretation:

The decrease in temperature when 2.25 moles of water initially at 310 K and 1650 bar pressure is brought to 1.30 bar in a reversible adiabatic process needs to be calculated

Concept Introduction:

A process is said to be adiabatic if there is not transfer of heat (q) between the system and surroundings i.e. q = 0

The entropy change (ΔS) associated with a process is the ratio of the heat transferred (q) and temperature (T)

  $\text{ΔS = }\frac{{\text{q}}_{\text{rev}}}{\text{T}}\text{------(1)}$

For a reversible adiabatic process, q = 0 therefore, ΔS = 0