Q6 / 150 · MHT CET 2026 (15 April Shift 1) · Chemistry
Calculate the compressibility factor of $1$ mole of a certain real gas if it occupies $0.4 \text{ dm}^3$ at $300$ K and $40$ atm. [R $= 0.082$ atm $\text{dm}^3 \text{K}^{-1}\text{mol}^{-1}$]
- A$0.45$
- Correct answerB$0.65$
- C$0.85$
- D$1.00$
Solution
The compressibility factor $Z$ is given by the formula:
$Z = \dfrac{PV}{nRT}$
Substituting the given values:
$Z = \dfrac{40 \times 0.4}{1 \times 0.082 \times 300}$
$Z = \dfrac{16}{24.6}$
$Z \approx 0.65$
Answer: $0.65$
$Z = \dfrac{PV}{nRT}$
Substituting the given values:
$Z = \dfrac{40 \times 0.4}{1 \times 0.082 \times 300}$
$Z = \dfrac{16}{24.6}$
$Z \approx 0.65$
Answer: $0.65$
Hence, the correct answer is option: 'B'.