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For one mole of a van der Waals gas when...

For one mole of a van der Waals gas when `b =0` and `T =30 K` the `PV vs1//V` plot is shown below The value of the van Waals constant a `("atm litre"^(2) mol^(-2))` is
.

A

`1.0`

B

`4.5`

C

`1.5`

D

`3.0`

Text Solution

Verified by Experts

The correct Answer is:
C
van der Waals' equation is :
`(P+(a)/(V^(2)))(V-b)=RT` (For 1 mole of gas)
Since `b =0`
`V(P+(a)/(V^(2))) = RT` or `PV = RT - a/V`
On comparing with `y = mx + C`
Intercept `(c) = RT`
Slope `(m) = -a`
Slope `= (y_(2)-y_(1))/(x_(2)-x_(1)) = (20.1-21.6)/(3-2)=-1.5`
`-a=-1.5` or `a=1.5`
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