At high temperature and low pressure, the Vander Waals equation is reduced to:

(p + \frac{a}{V_{m}^{2}}) (V_{m}) = R T

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{p V}_{m} = R T

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p (V_{m} - b) = R T

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(p + \frac{a}{V_{m}^{2}}) (V_{m} - b) = R T

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Solution

The correct option is A ${p V}_{m} = R T$
At high temperature and low pressure, $V$ is a larger comparison to $^{'} b^{'}$ and $(\frac{a}{V^{2}})$ is negligible in comparison to Pressure(P).
Hence, the equation reduces to $P V = R T$ .

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Similar questions

(p + \frac{a}{V_{m}^{2}}) (V_{m} - b) = R T

P (V_{m} - b) = R T e x p (- a / R T V_{m})

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

Z

P (V_{m} - b) = R T .

{(\frac{δ Z}{δ P})}_{T} is \frac{x b}{R T} .

x

mole

2

1

\begin{matrix} Column 1 & Column 2 A & At low pressure & i & P (V - b) = R T B & At high pressure & i i & (P + \frac{a}{V^{2}}) V = R T C & At low pressure and high temperature & i i i & P V = R T D & At room temperature and pressure & i v & (P + \frac{a}{V^{2}}) (V - b) = R T \end{matrix}

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