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QJEEM19Ch187e: JEE Main 2019 Chemistry - Class 11 JEEM19Ch187e Solution: Consider the following table : .tg…
Consider the following table :
a and b are vander Waals constants. The correct statement about the gases is :
| Gas | a/(k Pa dm6 mol-1) | b/(dm3 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 vander Waals constants. The correct statement about the gases is :
Multiple Choice Options
- A Gas C will occupy more volume than gas A; gas B will be more compressible than gas D
- B Gas C will occupy lesser volume than gas A; gas B will be more compressible than gas D
- C Gas C will occupy lesser volume than gas A; gas B will be lesser compressible than gas D
- D Gas C will occupy more volume than gas A; gas B will be lesser compressible than gas D
Step-by-Step Verified Solution
van der Walls equation of state is,
b = intermolecule volume of gases.
So when the value of 'b' is higher then the higher value of ‘b’ will occupy higher volume and according to van der Walls equation of state 'z' will be higher hence gas will be less compressible.
The value of 'a' is higher means intermolecule force of attraction is higher then the volume of gas will be lesser and according to van der Walls equation of state 'z' will be smaller hence gas will have higher compressibility.
b = intermolecule volume of gases.
So when the value of 'b' is higher then the higher value of ‘b’ will occupy higher volume and according to van der Walls equation of state 'z' will be higher hence gas will be less compressible.
The value of 'a' is higher means intermolecule force of attraction is higher then the volume of gas will be lesser and according to van der Walls equation of state 'z' will be smaller hence gas will have higher compressibility.