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Van der Waal's Equation

For one mole ...

Question

For one mole of a van der waal's gas when $b = 0 a n d T = 300 K, t h e P V v s, 1 / V p l o t$ . The value of the van der waal's constant $^{'} a^{'} (a t m . l i t r e^{2} m o l^{- 2})$ is

3.0

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1.5

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4.5

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1.0

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Solution

The correct option is B $1.5$
$(P + \frac{a}{V^{2}}) V = R T$

$P V + \frac{a}{V} = R T$

$P V = R T - \frac{a}{V}$

So, plot of $(P V) v s (\frac{1}{V})$ has slope - a.

$& t h e r e 4; s l o p e = \frac{20.1 - 24.6}{3} = - 1.5$

$& t h e r e 4; a = 1.5 a t m L^{2} m o l^{- 2}$

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Similar questions

b = 0

T = 300 K

P V

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a

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\frac{1}{V}

({atm. liter}^{2} {mol}^{- 2})

\frac{1}{V}

({atm. liter}^{2} {mol}^{- 2})

p V v s \frac{1}{V}

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

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