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The compressibility factor for definite ...

The compressibility factor for definite amount of van der Waals' gas at `0^(@)C` and `100 atm` is found to be `0.5`. Assuming the volume of gas molecules negligible, the van der Waals' constant `a` for gas is

A

`1.256 L^(2) mol^(-2) atm`

B

`0.256 L^(2) mol^(-2) atm`

C

`2.256 L^(2) mol^(-2) atm`

D

`0.0256 L^(2) mol^(-2) atm`

Text Solution

Verified by Experts

`Z=(PV)/n(RT)=0.5`
Now, `[P+(n^(2)a)/(V^(2))][V]=nRT` (`b` is negligible)
`PV^(2)-nRTV+n^(2)a=0`
`:. V=(nRT+-sqrt(n^(2)R^(2)T^(2)-4n^(2)axxb))/(2P)`
Since `V` is constant at given `P` and `T`, thus, discriminant is `0`.
`:. N^(2)R^(2)T^(2)=4n^(2)aP` or `a=(R^(2)T^(2))/(4P)`
`((0.0821)^(2)xx(273)^(2))/(4xx100)`
`=1.256 L mol^(-2) atm`
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The compression factor (compressibility factor) for 1 mol of a van der Waals gas at 0^(@)C and 100 atm pressure is found to be 0.5 . Assuming that the volume of a gas molecule is neligible, calculate the van der Waals constant a .

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