Chapter 3, Problem 3.60P

Interpretation Introduction

Interpretation:

To find the mass of $n$ -butane vapors in the tank which are in equilibrium with $\text{14}{\text{m}}^{\text{3}}$ of its $n$ -butane liquid in $\text{30}{\text{m}}^{\text{3}}$ tank at ${\text{25}}^{\text{o}}\text{C}$ . The vapor pressure of $n$ -butane at given temperature is $2.43\text{bar}$ .

Concept Introduction:

The molar volume of butane vapor in vessel is given as:

  $V=\frac{V_{vapor}}{n}$

The mass of $n$ -butanevapours present in tank is:

  $\begin{array}{l}m=n\times M\\ m=\frac{V_{vapor}}{V}\times M\end{array}$

Where $n$ is number of moles of butane vapor present in tank, $M$ is molecular weight of butane.

The $V$ is molar volume and $V_{vapor}$ is total volume of vapors in tank.

The molar volume of butane vapor in vessel is also calculated from generalized correlations for gases given by Pitzer at given conditions. The expression is:

$Z=1+B^0\frac{P_r}{T_r}+\omega B^1\frac{P_r}{T_r}$ where $PV=ZRT$........(1)