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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
van der Waals equations is `(P+(a)/(V^(2)))(V-b)=RT`
As given that b = 0,
`PV+(a)/(V)=RT rArr 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)=-1.5` . Thus, a = 1.5
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