Question

If at 200 K & 500 atm, density of $CH_4$ is 0.25 gm/ml then its compressibility factor (Z) is

• A. 1
• B. 2
• C. 0.5
• D. 0.25

Answer 1

Answer: (B)

2

Answer 2

The compressibility factor (Z) is calculated using the formula $Z = \frac{PM}{\rho RT}$. First, convert the density from g/ml to g/L: $0.25 \text{ g/ml} = 250 \text{ g/L}$. The molar mass of $CH_4$ is 16 g/mol. Substitute the given values: P = 500 atm, M = 16 g/mol, $\rho$ = 250 g/L, R = 0.08 L-atm/K-mol, and T = 200 K. $Z = \frac{(500 \text{ atm}) \times (16 \text{ g/mol})}{(250 \text{ g/L}) \times (0.08 \text{ L-atm/K-mol}) \times (200 \text{ K})} = \frac{8000}{4000} = 2$.