Question

The density of steam at ${100}^{o}C$ and $1\times {10}^5$ $Pa$ is $0.6kg{m}^{-3}$. The compressibility factor for steam is:

• A. 1
• B. 0.967
• C. 1.03
• D. 0.867

Answer 1

The correct option is B 0.967

As we know,
comprssibility factor ($Z$)
$Z=\frac{PV}{nRT}=\frac{PV\times M}{wRT}=\frac{P\times M}{\rho \times RT}=\frac{1\times {10}^5\times 18\times {10}^{-3}}{0.6\times 8.314\times 373}$
$=0.967$