Question

The compressibility factor Z for the gas at ${327}^{o}C$ and 60.375 atm pressure is :

• A. $1.226$
• B. $1$
• C. $0.226$
• D. $1.456$

Answer 1

The correct option is A $1.226$

Given,

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

Also as we know,

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

$M=\frac{2}{0.125-0.075}=40$ so molar mass of gas is 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 $40$ so gas is Ar.

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

Also, compressibility factor Z

$Z=\frac{PV}{nRT}=\frac{60.375\times 5}{5\times 0.0821\times 600}=1.226$