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Compressibility Factor

Compressibili...

Question

Compressibility factor for $C O_{2}$ at $400 K$ and $71.0 bar$ is $0.8697$ . Molar volume of $C O_{2}$ under these conditions is:

A

0.40 L

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B

2.24 L

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C

22.4 L

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D

19.5 L

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Solution

The correct option is A $0.40 L$
Given:
$Z = 0.8697, T = 400 K, P = 71 bar$
$Z = \frac{P V}{n R T} = \frac{P V_{m}}{R T}$
Molar volume of $C O_{2}, V_{m} = \frac{Z R T}{P}$
$V_{m} = \frac{0.8697 \times 0.0821 \times 400}{71} = 0.40 L$

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Compressibility Factor

Standard XII Chemistry

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