Chapter 6, Problem 6.9P

Interpretation Introduction

Interpretation:

Find the values of $Q$, $W$, $\Delta U$, $\Delta S$ and $\Delta H$ for one kilogram of water of volume $V_1=1\text{003}\frac{{\text{cm}}^{\text{3}}}{\text{kg}}$ in a piston-cylinder device at $\text{298}\text{.15 K}$ and $1\text{bar}$ is compressed in a mechanically reversible isothermal process to $1500\text{bar}$ given that $\beta ={\text{250×10}}^{\text{-6}}{\text{K}}^{\text{-1}}$ and $\kappa ={\text{45×10}}^{\text{-6}}{\text{bar}}^{\text{-1}}$

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

Concept Introduction:

For liquids or water as function of pressure and constant temperature in favor of volume expansivity and isothermal compressibility following equations are:

  $\frac{dH}{dP}=\left(1-\beta T\right)V_{avg}\approx \frac{\Delta H}{\Delta P}=\left(1-\beta T\right)V_{avg}$

  ${\frac{dS}{dP}}_T=-\beta V_{avg}\approx \frac{\Delta S}{\Delta P}=-\beta V_{avg}$

  ${\frac{dU}{dP}}_T=\left(\kappa P-\beta T\right)V_{avg}\approx {\frac{\Delta U}{\Delta P}}_T=\left(\kappa P-\beta T\right)V_{avg}$

  $dS=\frac{Q}{T}\approx \Delta S=\frac{Q}{T}$

  $W=\Delta U-Q$

  $V_{avg}=\frac{V_1+V_2}{2}$

And

  $\begin{array}{l}\kappa =\frac{-1}{V}{\frac{dV}{dP}}_T\\ V_2=V_1\exp (-\kappa (P_2-P_1))\end{array}$