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For one mole of a van der Waals' gas whe...

For one mole of a van der Waals' gas when `b=0` and `T=300K`, the `pV vs 1//V` plot is shown below. The value of the vander Waals' constant `a`(atm `L mol^(-2)`)

A

`1.0`

B

`4.5`

C

1.5

D

`3.0`

Text Solution

Verified by Experts

The correct Answer is:
C
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`
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