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Standard X

Chemistry

Ideal Gas Equation

The density o...

Question

The density of water vapour at $328.4$ atm and $800$ K is $135.0$ g $/ d m^{3}$ . Determine the molar volume, $V_{m}$ and the compression factor of water vapour.

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Solution

The molar volume of th vapour, $V_{m} = \frac{M_{1}}{ρ}$
$∴ V_{m} = \frac{18.09 g m o l^{- 1}}{135 g d m^{- 3}} = 0.134 d m^{3} m o l^{- 1}$
The compression factor, $Z = \frac{V_{m, r e a l}}{V_{m, i d e a l}} = \frac{V_{m, r e a l}}{R T / ρ}$
$= \frac{328.4 a t m \times 0.134 d m^{3} m o l^{- 1}}{0.0821 l i t a t m K^{-} m o l^{- 1} \times 800 K}$
$= 8.641$

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The density of water vapour at 328.4 atm and 800K is 135.0 g/dm3. Determine the molar volume, Vm and the compression factor of water vapour.

The molar volume of th vapour, Vm=M1ρ∴Vm=18.09gmol−1135gdm−3=0.134dm3mol−1The compression factor, Z=Vm,realVm,ideal=Vm,realRT/ρ =328.4atm×0.134dm3mol−10.0821litatmK−mol−1×800K =8.641

The density of water vapour at $328.4$ atm and $800$ K is $135.0$ g $/ d m^{3}$ . Determine the molar volume, $V_{m}$ and the compression factor of water vapour.