Question

A mixture containing 1.6 g of $O_2$, 1.4 g of $N_2$ and 0.4 g of $He$ occupies a volume of $10L$ at ${27}^{\circ }C$. Calculate the partial pressure of each component of the mixture.

• A. $P_{N_2}=0.500atm$
  $P_{O_2}=0.25atm$
  $P_{He}=0.25atm$
  
  
• B. $P_{N_2}=0.246atm$
  $P_{O_2}=0.246atm$
  $P_{He}=0.492atm$
  
  
• C. $P_{N_2}=0.123atm$
  $P_{O_2}=0.123atm$
  $P_{He}=0.246atm$
• D. $P_{N_2}=1atm$
  $P_{O_2}=1.123atm$
  $P_{He}=2.246atm$
  

Answer 1

The correct option is C

$P_{N_2}=0.123atm$
$P_{O_2}=0.123atm$
$P_{He}=0.246atm$

$P_{He}=0.246atm,P_{O_2}=P_{N_2}=0.123atm$

Given that $V=10L$ , T = ${27}^{\circ }C=300K$
The number of moles are :-

$n_{He}=\frac{0.4}{4}=0.1$
$n_{N_2}=\frac{1.4}{28}=0.05$
$n_{O_2}=\frac{1.6}{32}=0.05$

therefore , the total number of moles = 0.1+0.05+0.05 = 0.2

Using the ideal gas equation PV = nRT $\Rightarrow P\times 10=0.2\times 0.082\times 300\Rightarrow P=0.492atm$

As partial pressure = total pressure $\times$ mole fraction , we have

$P_{He}=0.492\times \frac{0.1}{0.2}$ = 0.246 atm

$P_{N_2}=0.492\times \frac{0.05}{0.2}=0.123atm$

$P_{O_2}=0.492\times \frac{0.05}{0.2}$ = 0.123 atm