Question

A mixture of $28g\;{N_2}$ , $8g\;He$ and $40g\;Ne$ has $20\;bar$ pressure. What is the partial pressure of each of these gases?

Answer 1

We are discussing the partial pressure of gases here. We should know what partial pressure is. Partial pressure of each gas is defined as the pressure exerted by each gas on the wall of the container carrying the gases.

Complete answer:
Total partial pressure exerted on the container carrying the gases is given by the sum of the partial pressure of each gas present in the container. Partial pressure is calculated by the formula
${{\text{P}}_{\text{T}}}\; = \;{{\text{P}}_{\text{A}}}\; + \;{{\text{P}}_{\text{B}}}\; + \;{{\text{P}}_{\text{C}}}$
where ${{\text{P}}_{\text{T}}}$ is the total partial pressure exerted by the mixture of three gases A, B and C on the container, ${{\text{P}}_{\text{A}}}$ is the partial pressure exerted by gas A, ${{\text{P}}_{\text{B}}}$ is the partial pressure exerted by gas B, ${{\text{P}}_{\text{C}}}$ is the partial pressure exerted by gas C. Partial pressure of each gas is given by
${{\text{P}}_{\text{A}}}\; = \;{{\text{x}}_{\text{A}}}{{\text{P}}_{\text{T}}}$
where ${{\text{x}}_{\text{A}}}$ is the mole fraction of the gas A
similarly,
${{\text{P}}_{\text{B}}}\; = \;{{\text{x}}_{\text{B}}}{{\text{P}}_{\text{T}}}$
${{\text{P}}_{\text{C}}}\; = \;{{\text{x}}_{\text{C}}}{{\text{P}}_{\text{T}}}$
where ${{\text{x}}_{\text{B}}}$ is the mole fraction of gas B, ${{\text{x}}_{\text{C}}}$ is the mole fraction of gas C
Now let’s see the values given to us in the question
weight of Nitrogen, ${{\text{N}}_2} = 28{\text{g}}$
weight of Helium, $He = 8{\text{g}}$
weight of Neon, $Ne = 20{\text{g}}$
Total partial pressure of the mixture, ${{\text{P}}_{\text{T}}} = 20\;bar$
let’s calculate number of moles of each gas then mole fraction of each gas
${{\text{n}}_{{{\text{N}}_2}}} = \dfrac{{28}}{{14}} = 2\;mol$
similarly for helium and neon
${{\text{n}}_{He}} = \dfrac{8}{4} = 2\;mol$
${{\text{n}}_{Ne}} = \dfrac{{40}}{{20}} = 2\;mol$
now we calculate mole fractions of gas A, B and C
${{\text{x}}_{\text{A}}} = \dfrac{{{{\text{x}}_{\text{A}}}}}{{{{\text{x}}_{\text{A}}} + {{\text{x}}_{\text{B}}} + {{\text{x}}_{\text{C}}}}}$
${{\text{x}}_{\text{A}}} = \dfrac{2}{{2 + 2 + 2}} = \dfrac{1}{3}$
similarly for gas B and C
${{\text{x}}_{\text{B}}} = \dfrac{2}{{2 + 2 + 2}} = \dfrac{1}{3}$
${{\text{x}}_{\text{C}}} = \dfrac{2}{{2 + 2 + 2}} = \dfrac{1}{3}$
Now as we can see the mole fraction of each gas is equal, so the total partial pressure exerted by the mixture will be equally divided into three gases. Also, according to formula, we can see
${{\text{P}}_{\text{A}}} = 20\; \times \;\dfrac{1}{3} = \dfrac{{20}}{3}\;bar$
similarly for gas B and C
${{\text{P}}_{\text{B}}} = 20\; \times \;\dfrac{1}{3} = \dfrac{{20}}{3}\;bar$
${{\text{P}}_{\text{C}}} = 20\; \times \;\dfrac{1}{3} = \dfrac{{20}}{3}\;bar$
So Partial Pressure for Nitrogen, Helium and Neon is same $\dfrac{{20}}{3}\;bar$

Note:
Calculating mole fraction of each gas carefully is important for the partial pressure calculation of the gas. If the mole fraction of gases in a mixture is equal then the total partial pressure will be equally distributed among the gases.