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Real Gases

For one mole ...

Question

For one mole of a van der Waal's gas when $b = 0$ and $T = 300 K$ , the $P V$ vs $1 / V$ plot is shown above.

The value of the van der Waal's constant $a$ (in atm $l i t r e^{2} m o l^{- 2})$ 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 C $1.5$
The van der Waals' equation for $1 m o l$ gas will be :

$[P + \frac{a}{V^{2}}] [V - b] = R T$

Since, $b = 0$

$∴ P V = R T - \frac{a}{V} (y = c + m x f o r m)$

When $P V$ is plotted against $(1 / V)$ , we get straight line with negative slope i.e., $(- a)$

$∴ S l o p e = \frac{20.1 - 24.6}{3}$

$= - 1.5$ .

So, a = 1.5

Therefore, the correct option is C.

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For one mole of a van der Waal's gas when b=0 and T=300 K, the PV vs 1/V plot is shown above. The value of the van der Waal's constant a (in atm litre2mol−2) is :

The correct option is C 1.5The van der Waals' equation for 1 mol gas will be :[P+aV2][V−b]=RTSince, b=0∴PV=RT−aV(y=c+mx form)When PV is plotted against (1/V), we get straight line with negative slope i.e., (−a)∴Slope=20.1−24.63=−1.5.So, a = 1.5Therefore, the correct option is C.

For one mole of a van der Waal's gas when $b = 0$ and $T = 300 K$ , the $P V$ vs $1 / V$ plot is shown above.

The value of the van der Waal's constant $a$ (in atm $l i t r e^{2} m o l^{- 2})$ is :