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Real Gas Equation

For one mole ...

Question

For one mole of a vander Waals’ gas when b = 0 and T = 300K, the PV vs. $\frac{1}{V}$ plot is shown below. The value of the van der Waals constant ‘a’ $(a t m l i t r e^{2} m o l^{- 2})$ is:

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4.5

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1.5

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Solution

The correct option is C 1.5
Van der Waals equation for 1 mol of real gas is $[P + \frac{a}{V^{2}}] [V - b] = R T$ Given that b = 0.
$∴ (P + \frac{a}{V^{2}}) (V) = R T ∴ P V = R T - \frac{a}{V}$
Following y = mx + c for the curve PV $v s \frac{1}{v}$ .
Slope = –a
$S l o p e = \frac{216 - 20.1}{2 - 3} = - 1.5$
$∴ a = 1.5$

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For one mole of a vander Waals’ gas when b = 0 and T = 300K, the PV vs. 1V plot is shown below. The value of the van der Waals constant ‘a’ (atm litre2mol−2) is:

The correct option is C 1.5Van der Waals equation for 1 mol of real gas is [P+aV2][V−b]=RT Given that b = 0. ∴(P+aV2)(V)=RT∴PV=RT−aV Following y = mx + c for the curve PV vs1v. Slope = –a Slope=216−20.12−3=−1.5 ∴a=1.5

For one mole of a vander Waals’ gas when b = 0 and T = 300K, the PV vs. $\frac{1}{V}$ plot is shown below. The value of the van der Waals constant ‘a’ $(a t m l i t r e^{2} m o l^{- 2})$ is: