Consider the following table :
Gas \(a/(\mathrm{kPa\ dm^6\ mol^{-1}})\) \(b/(\mathrm{dm^3\ mol^{-1}})\) A 642.32 0.05196 B 155.21 0.04136 C 431.91 0.05196 D 155.21 0.4382
\(a\) and \(b\) are van der Waals constants. The correct statement about the gases is :
| Gas | \(a/(\mathrm{kPa\ dm^6\ mol^{-1}})\) | \(b/(\mathrm{dm^3\ mol^{-1}})\) |
| A | 642.32 | 0.05196 |
| B | 155.21 | 0.04136 |
| C | 431.91 | 0.05196 |
| D | 155.21 | 0.4382 |
Show Hint
- Gas C will occupy more volume than gas A; gas B will be more compressible than gas D
- Gas C will occupy lesser volume than gas A; gas B will be more compressible than gas D
- Gas C will occupy lesser volume than gas A; gas B will be lesser compressible than gas D
- Gas C will occupy more volume than gas A; gas B will be lesser compressible than gas D
The Correct Option is A
Solution and Explanation
Gas B and gas D have the same \(a\) (155.21), but \(b_B = 0.04136 < b_D = 0.4382\). A larger \(b\) means a larger repulsive (excluded-volume) contribution, so \(Z = \frac{PV}{RT}\) is greater and the gas is harder to compress. So gas D is less compressible and gas B is more compressible than gas D.
Hence: gas C occupies more volume than A; B is more compressible than D.
Answer: (1) Gas C will occupy more volume than gas A; gas B will be more compressible than gas D
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