For one mole of a Van der Waals gas when b - 0 and T - 300 K- the PV vs 1V plot is shown above- The value of the Van der Waals constant a atm-L2-mol-2 is

The correct option is C 1.5Vanderwaal's equation for 1 mole of gas is #160; (P+ a/V^2)(V b)=RT But, b=0 ......(given) (P+ a/V^2)(V)=RT ∴ PV ...

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Standard XII

Chemistry

Real Gases

For one mole ...

Question

For one mole of a Van der Waals gas when b = 0 and T = 300 K, the $P V$ vs $\frac{1}{V}$ plot is shown above. The value of the Van der Waals constant a $(a t m . L^{2} . m o l^{- 2})$ 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
Vanderwaal's equation for 1 mole of gas is
$(P + \frac{a}{V^{2}}) (V - b) = R T$

But, $b = 0$ ......(given)

$(P + \frac{a}{V^{2}}) (V) = R T$

$∴ P V = - a \times \frac{1}{V} + R T$

We know the basic general equation of straight line, $y = m x + c$

Slope $= t a n (π - θ) = - t a n θ = - a$

So, $t a n θ = a = \frac{21.6 - 20.1}{3 - 2} = 1.5$

Option C is correct.

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

The correct option is C 1.5Vanderwaal's equation for 1 mole of gas is (P+aV2)(V−b)=RTBut, b=0......(given)(P+aV2)(V)=RT∴PV=−a×1V+RTWe know the basic general equation of straight line, y=mx+cSlope =tan(π−θ)=−tan θ=−aSo, tanθ=a=21.6−20.13−2=1.5Option C is correct.

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