Question

The density of steam at ${27}^{o}C$ and $8.314\times {10}^4$ pascal is 0.8 Kg $m^3$. The compressibility factor would be:

• A. 0.75
• B. 1
• C. 0.88
• D. 1.1

Answer 1

Answer: (A)

0.75

The compressibility factor is defined as,

$Z=\frac{PV}{nRT}=\frac{PV}{\frac{w}{M}RT}$

$=\frac{PM}{dRT}=\frac{8.314\times {10}^4\times 18\times {10}^{-3}}{0.8\times 8.314\times 300}$

$=\frac{18\times 10}{300\times 0.8}=\frac{3}{4}=0.75$

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