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Standard XII

Chemistry

Surface Tension

The density o...

Question

The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg $m^{- 3}$ . The vapour effuses through a small hole at a rate 1.33 times faster than oxygen under the same condition. Determine compressibility factor? ( as an integer only)

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Solution

$\frac{r_{(v)}}{r (O_{2})} = \sqrt{\frac{M_{(O_{2})}}{M_{(v)}}}$

$∴ 1.33 = \sqrt{\frac{32}{M_{(v)}}}$

$M_{(v)} = 18.1 g m o l^{- 1}$

so integer value is 18.

Molar volume $(¯ ¯¯ ¯ V) = \frac{Molar mass}{Density of 1 mole}$

$= \frac{18.1 \times 10^{- 3}}{0.36}$

$= 50.26 \times 10^{- 3} m^{3}$

Compression factor $(Z) = \frac{P ¯ ¯¯ ¯ V}{R T} = \frac{101325 \times 50.25 \times 10^{- 3}}{8.314 \times 500}$

$(P = 101325 N m^{- 2} = 1 a t m)$

$= 1.225$

So, the nearest integer is $1$ .

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The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg m−3. The vapour effuses through a small hole at a rate 1.33 times faster than oxygen under the same condition. Determine compressibility factor? ( as an integer only)

r(v)r(O2)=√M(O2)M(v)∴1.33=√32M(v)M(v)=18.1gmol−1so integer value is 18.Molar volume (¯¯¯¯V)=Molar massDensity of 1 mole=18.1×10−30.36=50.26×10−3m3Compression factor (Z)=P¯¯¯¯VRT=101325×50.25×10−38.314×500(P=101325Nm−2=1atm)=1.225So, the nearest integer is 1.

The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg $m^{- 3}$ . The vapour effuses through a small hole at a rate 1.33 times faster than oxygen under the same condition. Determine compressibility factor? ( as an integer only)