Chapter 6, Problem 6.8P

Interpretation Introduction

Interpretation:

Calculate the change in temperature and entropy $\Delta T$ and $\Delta S$ for the liquid isobutane at $360\text{K}$ if it is throttled through a value from initial state of $3\text{60}\text{K}$ and $4000\text{kPa}$ to $\text{2000 kPa}$ given that $C_P=2.78\frac{\text{J}}{\text{g}\text{C}}$ . The values of $V^l$ and $\beta$ can be found out through generalized correlations for liquids:

$V^{sat}=V_c{Z}_c{}^{{(1-T_r)}^{2/7}}$............(1)

Concept Introduction:

For incompressible liquids which is not nearby its critical point and pressure has little effect on properties of liquids, following correlation for change in entropy $\Delta S$ is used:

  $dS=C_P\frac{dT}{T}-\beta VdP$............(2)

And

  $\Delta S=\frac{-V^l(P_2-P_1)}{T}$............(3)