Calculate the compressibility factor for $C O_{2}$ , it one mole of it occupies 0.4 L at 300K and 40 atm.

0.85

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0.65

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0.99

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Solution

The correct option is B 0.65
Compressibility Factor is also often defined as “the ratio of molar volume of gas to that of an ideal gas.” It is expressed as
Compressibility (Z) $= \frac{P V}{R T}$
Pressure, P = 40 atm, V= 0.4 L, R=0.082 L.atm. $K^{- 1} m o l^{- 1}$ , T= 300 K
$= \frac{40 \times 0.4}{0.821 \times 300}$ = $\frac{16}{246.3} = 0.65$

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