Question

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

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

Answer 1

The correct option is B $\frac{50}{41}$

The expression for the compressibility factor $z$ is $z=\frac{PV}{nRT}=\frac{PM}{\rho RT}$.

The density is $0.36kg/m^3$. It is also equal to $0.36g/L$.

Substitute values in the above expression:

$z=\frac{1\times 18}{0.36\times 0.082\times 500}=50/41$.