Question

A gaseous mixture of ${{\text{N}}_2}$ and gas A at 30${}^0{\text{C}}$ has a total pressure of 6 atm contains 30 mol percent of ${{\text{N}}_2}$. What is the partial pressure of A?
${\text{A}}{\text{.}}$ 4.2 atm
${\text{B}}{\text{.}}$ 1.8 atm
${\text{C}}{\text{.}}$ 3 atm
${\text{D}}{\text{.}}$ 2 atm

Answer 1

Hint: Here, we will proceed by finding the mole fraction corresponding to gas ${{\text{N}}_2}$ which will help in finding the mole fraction of gas A. Then, we will be using the Dalton’s Law of partial pressure in terms of mole fraction.
Formulas Used: $\sum {{{\text{X}}_{{\text{constituents}}}} = 1} $ and ${{\text{P}}_{\text{A}}} = {{\text{X}}_{\text{A}}} \times {{\text{P}}_{\text{T}}}$.

Complete answer:
Given, Temperature of a gaseous mixture of ${{\text{N}}_2}$ and gas A = 30${}^0{\text{C}}$
Total pressure of gaseous mixture, ${{\text{P}}_{\text{T}}}$ = 6 atm
Mol percent of ${{\text{N}}_2}$ = 30%
Mole fraction of ${{\text{N}}_2}$ can be given as under
Mole fraction of ${{\text{N}}_2}$, ${{\text{X}}_{{{\text{N}}_2}}}$ = $\dfrac{{{\text{Mol percent of }}{{\text{N}}_2}}}{{100}} = \dfrac{{30}}{{100}} = 0.3$
For any mixture, the sum of the mole fraction of the individual constituents is always equal to 1
i.e., $\sum {{{\text{X}}_{{\text{constituents}}}} = 1} $
Let the mole fraction of gas A in the gaseous mixture be ${{\text{X}}_{\text{A}}}$
The given gaseous mixture contains only ${{\text{N}}_2}$ and gas A, so we can say that the sum of the mole fraction of ${{\text{N}}_2}$ and gas A will always be equal to 1
i.e., $
  {{\text{X}}_{{{\text{N}}_2}}} + {{\text{X}}_{\text{A}}} = 1 \\
   \Rightarrow 0.3 + {{\text{X}}_{\text{A}}} = 1 \\
   \Rightarrow {{\text{X}}_{\text{A}}} = 1 - 0.3 = 0.7 \\
 $
According to Dalton’s Law of partial pressure in terms of mole fraction, we have
${{\text{P}}_{\text{A}}} = {{\text{X}}_{\text{A}}} \times {{\text{P}}_{\text{T}}}{\text{ }} \to {\text{(1)}}$ where ${{\text{P}}_{\text{A}}}$ denotes the partial pressure of gas A, ${{\text{X}}_{\text{A}}}$ denotes the mole fraction of gas A and ${{\text{P}}_{\text{T}}}$ denotes the total pressure of the gaseous mixture
By substituting the values of ${{\text{X}}_{\text{A}}}$ and ${{\text{P}}_{\text{T}}}$ in the formula given equation (1), we get
\[
   \Rightarrow {{\text{P}}_{\text{A}}} = {\text{0}}{\text{.7}} \times {\text{6}} \\
   \Rightarrow {{\text{P}}_{\text{A}}} = 4.2{\text{ atm}} \\
 \]
Therefore, the partial pressure of A in the gaseous mixture is 4.2 atm.

Hence, option A is correct.

Note:
The statement of Dalton’s Law of partial pressure is “The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases” which can be represented in mathematical form as ${{\text{P}}_{{\text{Total}}}} = {{\text{P}}_{{\text{Gas A}}}} + {{\text{P}}_{{\text{Gas B}}}} + {{\text{P}}_{{\text{Gas C}}}} + ....$. For given two gases, it will become ${{\text{P}}_{\text{T}}} = {{\text{P}}_{\text{A}}} + {{\text{P}}_{{{\text{N}}_2}}}$ which can be easily reduced to obtain ${{\text{P}}_{\text{A}}} = {{\text{X}}_{\text{A}}} \times {{\text{P}}_{\text{T}}}$.