If compressibility factor of real gas is 1.05 at STP. What is molar volume of real gas?
- $22 \cdot 40 \mathrm{dm}^3$
- $21 \cdot 33 \mathrm{dm}^3$
- $23.52 \mathrm{dm}^3$
- $24.52 \mathrm{dm}^3$
Solution
The compressibility factor $Z$ for a real gas is defined as $Z = \frac{V_{m,\text{real}}}{V_{m,\text{ideal}}}$.
At STP, the molar volume of an ideal gas is known to be $22.4\ \text{dm}^3$.
Given $Z = 1.05$, the molar volume of the real gas becomes $V_{m,\text{real}} = Z \times V_{m,\text{ideal}} = 1.05 \times 22.4 = 23.52\ \text{dm}^3$.
Among the options provided, both C and D state $23.52\ \text{dm}^3$, representing the correct value.
Asked in: MHT CET 2025 (20 April Shift 1)