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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 = 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

A

`1.0`

B

`4.5`

C

`1.5`

D

`3.0`

Text Solution

Verified by Experts

The correct Answer is:
C
`(P+(a)/(V^(2)) (V) = RT`
`PV +a//V RT`
`PV = RT -a(v)`
`y = RT - a(x)`
So, slope `=a =(21.6 -20.1)/(3-2) = (1.5)/(1) = 1.5`
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