Question

A $10$ litre flask contains $0.2$ mole of methane, $0.3mole$ of hydrogen and $0.4$ mole of nitrogen at ${25}^{\circ }C$. What is the partial pressure of each component and what is the pressure inside the flask?

Answer 1

Using ideal gas equation,

$P=\frac{nRT}{V}$
Partial pressure of methane $=\frac{0.2\times 0.0821\times 298}{10}=0.489atm$
Partial pressure of hydrogen $=\frac{0.3\times 0.0821\times 298}{10}=0.734atm$
Partial pressure of nitrogen $=\frac{0.4\times 0.0821\times 298}{10}=0.979atm$
Total pressure $=(0.489+0.734+0.979)atm$
$=2.202atm$
Alternative solution: Total number of moles
$=0.2+0.3+0.4=0.9$
Let the total pressure be $P$.
We know that,
$P=\frac{\text{total number of moles}}{V}\cdot RT$
$V=10litre,R=0.0821Latm{K}^{-1},T=(25+273)=298K$
$P=\frac{0.9}{10}\times 0.0821\times 298=2.20atm$
Partial pressure of $C{H}_4=$ Mole fraction of methane $\times$ total pressure
$=\frac{0.2}{0.9}\times 2.20=0.489atm$
Partial pressure of $H_2=$ Mole fraction of $H_2\times$ total pressure
$=\frac{0.3}{0.9}\times 2.20=0.733atm$
Partial pressure of $N_2=$ Mole fraction of $N_2\times$ total pressure
$=\frac{0.4}{0.9}\times 2.20=0.978atm$