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Q.
The reduced volume and reduced temperature of a gas are $10.2$ and $0.7$ respectively. If its critical pressure is $4.25\, atm$ then its pressure will be
AMUAMU 2015States of Matter
Solution:
Given that, Reduced volume, $\phi=10.2$
Reduced temperature, $\theta=0.7$ Critical
pressure, $p_{c}=4.25\, atm , p =$ ?
According to reduced equation of state
$\left(\pi+\frac{3}{\phi^{2}}\right)(3 \phi-1)$
$=8 \theta\left(\pi+\frac{3}{10.2 \times 10.2}\right)(3 \times 10.2-1)$
$=8 \times 0.7 \left(\pi+\frac{3}{104.04}\right)(30.6-1)$
$=5.6(\pi+0.0288)(29.6)=5.6 \pi+0.0288$
$=\frac{5.6}{29.6}=0.1892 \pi=0.1892-0.0288=0.1604$
Now, we know that $\pi=\frac{p}{p_{c}}$
$\therefore p=\pi \times p_{c} p=0.1604 \times 4.25=0.6816\, atm$