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Standard Form of a Hyperbola

For one mole ...

Question

For one mole of a van der Waal's gas when b = 0 and T = 300 K, the PV vs $\frac{1}{V}$ plot is shown below. The value of the van der Waal's constant a $({atm. liter}^{2} {mol}^{- 2})$ is:

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1.5

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4.5

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Solution

The correct option is B 1.5
As $b = 0$ , Van der waals equation for one mole of gas will be:
$(P + \frac{a}{V^{2}}) (V) = R T$
$P V + (\frac{a}{V}) = R T$
$P V = R T - (\frac{a}{V}) . . . (1)$
Now ,in the graph , we have y-axis as PV and x-axis as $(\frac{1}{V}) .$
Comparing Equation (1), with y = mx + c;
We obtain, slope $(m) = - a = \frac{[y_{2} - y_{1}]}{x_{2} - x_{1}} = \frac{[21.6 - 20.1]}{[2.0 - 3.0]}$
Slope (m) $= - a = - 1.5$
Hence $a = 1.5$

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Standard Form of a Hyperbola

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For one mole of a van der Waal's gas when b = 0 and T = 300 K, the PV vs 1V plot is shown below. The value of the van der Waal's constant a (atm. liter2mol−2) is:

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