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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

B

4.5

C

1.5

D

3

Text Solution

Verified by Experts

The correct Answer is:
C
For one mole of a real gas, van der Waals equation is
`(p+(a)/(V^(2)))(V-b)=RT`
when b = 0
`(p+(a)/(V^(2)))(V)=RT`
`"or "pV+(a)/(V)=RT`
`pV=RT-(a)/(V)`
Therefore, pV vs 1/V graph is a straight line having slope `=-a`
Slope of the given line `=-a=(y_(2)-y_(1))/(x_(2)-x_(1))`
`"or "-a=(21.6-20.1)/(2-3)`
`=-1.5" atm L"^(2)" mol"^(-2)`
`"or "="1.5 atm L"^(2)" mol"^(-2)`
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