For one mole of a Van der Waals gas when b - 0 and T - 300 K- then -pV vs- 1-V-plot is shown in figure- The value of the Van der Waals constant -a- -atm-L-2-mol-2-is- 3-0-1-0-4-5-1-5-

For one mole ...

Question

For one mole of a Van der Waals gas when b = 0 and T = 300 K, then pV vs. $\frac{1}{V}$ plot is shown in figure. The value of the Van der Waals constant "a" $(a t m L^{2} m o l^{- 2}$ )is___.

1.0

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4.5

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1.5

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3.0

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Solution

The correct option is D
1.5

$⟮ p + \frac{a}{V^{2}} ⟯ ⟮ V - b ⟯ = R T$

But it is given that b = 0, so , the equation reduces to

$⟮ p + \frac{a}{V^{2}} ⟯ V = R T \Rightarrow p V = \frac{a}{V} + R T$

Comparing it with a straight line equation , we get slope as -a. Calculating the slope , we get

$\frac{24.6 - 20.1}{3.0 - 0} = 1.5 \Rightarrow a = 1.5$

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For one mole of a Van der Waals gas when b = 0 and T = 300 K, then pV vs. $\frac{1}{V}$ plot is shown in figure. The value of the Van der Waals constant "a" $(a t m L^{2} m o l^{- 2}$ )is___.

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For one mole of a Van der Waals gas when b = 0 and T = 300 K, then pV vs. $\frac{1}{V}$ plot is shown in figure. The value of the Van der Waals constant "a" $(a t m L^{2} m o l^{- 2}$ )is___.

For one mole of a Van der Waals gas when b = 0 and T = 300 K, then pV vs. 1V plot is shown in figure. The value of the Van der Waals constant "a" (atm L2 mol−2)is___.

The correct option is D 1.5 ⟮ p+aV2⟯⟮V−b⟯=RT But it is given that b = 0, so , the equation reduces to ⟮ p+aV2⟯V=RT⇒pV=aV+RT Comparing it with a straight line equation , we get slope as -a. Calculating the slope , we get 24.6−20.13.0−0=1.5⇒a=1.5

For one mole of a Van der Waals gas when b = 0 and T = 300 K, then pV vs. $\frac{1}{V}$ plot is shown in figure. The value of the Van der Waals constant "a" $(a t m L^{2} m o l^{- 2}$ )is___.