For one mole of a van der Waals gas when b = 0 and T = 300 K , the PV vs 1/V plot is shown below . The value of the van der Waals constant a (atm `"litre"^(2) mol^(-2)`) is

Options

A. (a) `1.0`
B. (b) `4.5`
C. (c) 1.5
D. (d) `3.0`

Answer

C

Explanation

The van der Waals’ equation of state is `(p + (n^(2)a)/(V^(2))) (V-nb) = nRT`
For one mole and when b = 0, the above equation condenses to `(p + (a)/(V^(2)))V = RT`
`rArr pV = RT - (a)/(V)` …(i)
Eq. (i) is a straight equation between pV and `(1)/(V)` whose slope is `-a` . Equating with slope of the straight line given in the graph. `-a = (20.1 - 21.6)/(3-2) = -1.5 " " a = 1.5`