The compressibility factor $(Z)$ of one mole of a van der Waals' gas of negligible $^{'} a^{'}$ value is:

1

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\frac{b p}{R T}

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1 + \frac{b p}{R T}

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1 - \frac{b p}{R T}

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Solution

The correct option is C $1 + \frac{b p}{R T}$
van der Waal's equation is

$(p + \frac{a}{V^{2}}) (V - b) = R T$ (for 1 mole)

If $a$ is negligible, $p + \frac{a}{V^{2}} \approx p$

$p (V - b) = R T \Rightarrow p V - p b = R T$

or $\frac{p V}{R T} - \frac{p b}{R T} = 1 \Rightarrow \frac{p V}{R T} = 1 + \frac{b p}{R T}$

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Similar questions

R T = 20 L a t m m o l^{- 1}

a = 1000 a t m L^{2} m o l^{- 2}

a

Z = 1 + \frac{b P}{R T}

\frac{a}{V^{2}}

Z

a

Z

b

Z = 1 + \frac{b P}{R T} - \frac{a}{R T V} + \frac{a b}{R T V^{2}}

z = 1 + \frac{B P}{R T}

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