Question

In an experiment at constant temperature, a real gas X shows its compressibility factors equal to 1.6 & 1.8 at pressures 600 bar & ‘P’ bar respectively. What is the approximate value of ‘P’ in bar -

• A. 675
• B. 533.33
• C. 711.11
• D. 800

Answer 1

The correct option is D 800

Real gas follows the eqaution - $(P+\frac{a{n}^2}{V^2})(V-nb)$ = nRT
At higher Pressure the pressure correction term can be neglected -
$\Rightarrow$ P(V - nb) = nRT
$\Rightarrow$ $\frac{PV}{nRT}$ = 1 + $\frac{Pb}{RT}$
$\Rightarrow$ Z - 1 = $\frac{Pb}{RT}$
Thus, $\frac{Z_1-1}{Z_2-1}=\frac{P_1}{P}$
$\Rightarrow$ $\frac{1.6-1}{1.8-1}=\frac{600bar}{P}$ $\Rightarrow$ P = 800 bar
Hence correct answer is option (d)