Question

The reduced volume and reduced temperature of a gas are $10.2$ and $0.7$ respectively. If its critical pressure is $4.25atm$ then its pressure will be

• A. 0.6816 atm
• B. 0.6618 atm
• C. 0.8616 atm
• D. zero

Answer 1

Answer: (A)

Given that, Reduced volume, $\varphi =10.2$
Reduced temperature, $\theta =0.7$ Critical
pressure, $p_c=4.25atm,p=$ ?
According to reduced equation of state
$\left(\pi +\frac{3}{{\varphi }^2}\right)(3\varphi -1)$
$=8\theta \left(\pi +\frac{3}{10.2\times 10.2}\right)(3\times 10.2-1)$
$=8\times 0.7\left(\pi +\frac{3}{104.04}\right)(30.6-1)$
$=5.6(\pi +0.0288)(29.6)=5.6\pi +0.0288$
$=\frac{5.6}{29.6}=0.1892\pi =0.1892-0.0288=0.1604$
Now, we know that $\pi =\frac{p}{p_c}$
$\therefore p=\pi \times p_{c}p=0.1604\times 4.25=0.6816atm$