Question

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. \(0.45\)
• B. \(0.65\)
• C. \(0.85\)
• D. \(1.00\)

Hint:

Compressibility factor is Z = PV / nRT, which compares the real molar volume with the ideal one.

Answer 1

The Correct Option is B

Step 1: Ideal volume.
For an ideal gas: $V_{ideal} = \dfrac{nRT}{P} = \dfrac{24.6}{40} = 0.615$ dm$^3$.

Step 2: Ratio.
$Z = \dfrac{V_{real}}{V_{ideal}} = \dfrac{0.4}{0.615} = 0.65$.

Final Answer:
Option (B). \[ \boxed{Z = 0.65} \]