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Van der Waal's Equation

For one mole ...

Question

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

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 C $1.5$
We know,
Vander Waal's equation, for one mole of the gas,
$(p + \frac{a}{V^{2}}) (V - b) = R T$
Since,
$b = 0$
$(P + \frac{a}{V^{2}}) (V) = R T$
$P V + a / V = R T$
$P V = R T - a (v)$
This is an equation of a straight line.
$y = c - m (x)$

slope - m = - a

So slope $- a = \frac{20.1 - 21.6}{3 - 2} = \frac{1.5}{1} = - 1.5$
$a = 1.5$

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

The correct option is C 1.5We know, Vander Waal's equation, for one mole of the gas, (p+aV2)(V−b)=RT Since, b=0 (P+aV2)(V)=RT PV+a/V=RT PV=RT−a(v) This is an equation of a straight line. y=c−m(x) slope - m = - a So slope −a=20.1−21.63−2=1.51=−1.5 a=1.5

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