The deviation from the ideal gas behaviour of a gas can be expressed as

Z = \frac{p}{V R T}

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Z = \frac{p V}{n R T}

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Z = \frac{n R T}{p V}

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Z = \frac{V R}{p T}

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Solution

The correct option is A $Z = \frac{p V}{n R T}$
According to ideal gas equation,
$p V = n R T$
$n =$ number of moles of gas
Then $\frac{p V}{n R T} = 1$
Therefore compressibility factor
$Z = \frac{p V}{n R T} = 1$
For an ideal gas $Z = 1$ and for real gas, $Z$ may be either greater than one or less than one.

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