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Standard XIII

Chemistry

Compressibility Factor

The density o...

Question

The density of vapours of substance at 1 atm and $600 K$ is $0.32 k g / m^{3}$ . Vapours effuse through a small orifice at a rate of $1.414$ times faster than $O_{2}$ at same conditions. The $Z$ of the vapours is:
$R = \frac{1}{12} l a t m / m o l K; \frac{25}{3} J / m o l K$

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Solution

When the gas effuses, $velocity \propto \sqrt{\frac{1}{M}}$
$\frac{V_{x}}{V_{O_{2}}} \propto \sqrt{\frac{M_{O_{2}}}{M_{x}}}$
$\Rightarrow \frac{V_{x}}{V_{O_{2}}} = \sqrt{\frac{32}{x}} = 1.414$
$\Rightarrow \sqrt{\frac{32}{x}} = \sqrt{2}$
$\Rightarrow x = 16 g / m o l$
Molar volume of the gas, $V_{m} = \frac{16 g}{m o l} \times \frac{1 k g}{1000 g} \times \frac{m^{3}}{0.32 k g}$
$= 50 \times 10^{- 3} m^{3} = 50 L / m o l$

Compressibility factor , $Z = \frac{P V_{m}}{n R T}$
$Z = \frac{1 a t m \times 50 L / m o l}{\frac{1}{12} L a t m / m o l K \times 600 K} = 1$

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The density of vapours of substance at 1 atm and 600 K is 0.32 kg/m3. Vapours effuse through a small orifice at a rate of 1.414 times faster than O2 at same conditions. The Z of the vapours is: R=112 l atm/mol K ; 253 J/mol K

When the gas effuses, velocity ∝√1M VxVO2∝√MO2Mx ⇒VxVO2=√32x=1.414 ⇒√32x=√2 ⇒x=16 g/mol Molar volume of the gas, Vm=16 gmol×1 kg1000 g×m30.32 kg =50×10−3 m3=50L/mol Compressibility factor , Z=PVmnRT Z=1 atm×50 L/mol112 L atm/mol K×600 K=1

The density of vapours of substance at 1 atm and $600 K$ is $0.32 k g / m^{3}$ . Vapours effuse through a small orifice at a rate of $1.414$ times faster than $O_{2}$ at same conditions. The $Z$ of the vapours is:
$R = \frac{1}{12} l a t m / m o l K; \frac{25}{3} J / m o l K$