Question

The density of steam at $27{}^{\circ }C$ and $8.3174\times {10}^4$ pascal is $0.8kg{m}^{-3}$. The compressibility factor would be

• A. 0.75
• B. 1
• C. 0.88
• D. 1.1

Answer 1

The correct option is A 0.75

The compressibility factor is given as

$Z=\frac{PV}{nRT}$

The density of steam is $d$$=$$\frac{w}{V}$

$Z=\frac{PV}{nRT}=\frac{PV}{\frac{w}{M}RT}=\frac{PM}{dRT}=\frac{8.314\times {10}^4\times 18\times {10}^{-3}}{0.8\times 8.314\times 300}=\frac{18\times 10}{300\times 0.8}=\frac{3}{4}=0.75$