First slide

Gaseous state

Question

Compressibility factor for CO₂ at 400 K and 71.0 bars is 0.8697. Molar volume of CO₂ under these conditions is

Moderate

A

22.4 L
B

2.24 L
C

0.41 L
D

19. 5 L

Solution

$Z = \frac{pV}{nRT} = 0 .8697 ∴ V_{m} = \frac{0 .8697 \times nRT}{p} = \frac{0 .8697 \times 1 \times 0 .083 \times 400}{71} = 0 .41 L (R = 0 .083 bar {dm}^{3} K^{- 1} {mol}^{- 1})$

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