Calculate the compressibility factor for $S O_{2}$ , if $1 m o l e$ of it occupies $0.35 l i t r e$ at $300 K$ and $50 a t m$ pressure. Comment on the result.

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Solution

Compressibility factor = $Z = \frac{P V}{n R T} = \frac{50 \times 0.35}{1 \times 0.0821 \times 300} = 0.711$
Since, $Z < 1$ ; the gas $S O_{2}$ will be more compressible than ideal gas.

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1 mole of $S O_{2}$ occupies a volume of $350 m l$ at $300 K$ and $50 a t m$ pressure. Calculate the compressibility factor of the gas.

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