Question

If the gas $X$ obeys van der Waal's equation and if the value of $a=1.2atmlitr{e}^{2}mo{l}^{-1}$ and $b=0.2litremo{l}^{-1}$, then pressure $P$ of gas at ${327}^{o}C$ is :

• A. $30.18$ atm
• B. $60.375$ atm
• C. $120.75$ atm
• D. $90.55$ atm

Answer 1

Answer: (B)

$60.375$ atm

Given,
$C_p=0.125cal/g,C_v=0.075cal/g$

Also as we know,

$C_p-C_v=\frac{R}{M}$

$M=\frac{2}{0.125-0.075}=40$

Also, $\frac{C_p}{C_v}=\frac{0.125}{0.075}=\frac{5}{3}=1.66$, so gas is monoatomic, also molar mass of gas is $40$ g/mol.

Thus, gas is Ar.

According to vander waal's equation :

$[P+\frac{n^{2}a}{V^2}][V-nb]=nRT$

Also, $n=\frac{w}{M}=\frac{200}{40}=5$

$[P+25\times \frac{1.2}{25}][5-5\times 0.2]=5\times 0.0821\times 600$

$P=60.375$ atm.