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
.

Options

A. `1 .0`
B. `4.5`
C. `1.5`
D. `3.0`

Explanation

van der Waals' equation is
`[P+(a)/(V^(2))][V-b]=RT`
`:. B =0`
`:. PV + (a)/(V) =RT`
or `PV =RT - (a)/(V)`
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)`
Thus `a =+ 1.5` .