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 L mol $^{- 2}$ )

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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 the straight line, $y = m x + c$

Slope $= t a n (π - θ) = - a$

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

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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 L mol−2)

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 the straight line, y=mx+cSlope =tan(π−θ)=−aSo, tanθ=a=21.6−20.13−2=1.5

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 L mol $^{- 2}$ )