Question

If density of vapours of a substance (molar mass 18 gm/mole) at 1 atm and 500 K is 0.36 $kg{m}^{-3}$, the value of Z for the vapours is:

[Take $R=0.082Latmmol{e}^{-1}{K}^{-1}]$

• A. $\frac{41}{50}$
• B. $\frac{50}{41}$
• C. $1.1$
• D. $0.9$

Answer 1

Answer: (B)

$\frac{50}{41}$

The expression for the actual volume of the gas due to non ideal behaviour is $V_{real}=\frac{\text{Molar mass}}{density}=\frac{18}{0.36}=50$.

The expression for the volume of the gas due to ideal behaviour is $V_{ideal}=\frac{nRT}{P}=\frac{1\times 0.082\times 500}{1}$.

The compressibility factor Z is the ratio of the actual volume of the gas due to non-ideal behaviour to the volume of the gas due to ideal behaviour.

Thus, $Z=\frac{V_{real}}{V_{ideal}}=\frac{50}{0.082\times 500}=\frac{50}{41}$

Thus the value of the compressibility factor is $\frac{50}{41}$.

Hence, the correct answer is option $\text{B}$.