Question

The pressure of a $1:4$ mixture by moles of dihydrogen and dioxygen enclosed in a vessel is one bar. What will be the partial pressure of dioxygen?

• A. $0.8\times {10}^4\text{bar}$
• B. $0.008N{m}^{-2}$
• C. $8\times {10}^{4}N{m}^{-2}$
• D. $0.08\text{bar}$

Answer 1

The correct option is C $8\times {10}^{4}N{m}^{-2}$

Let the number of moles of dihydrogen and dioxygen be 1 and 4.
$\therefore$ Mole fraction of $O_2=\frac{4}{1+4}=\frac{4}{5}$
So, Partial pressure of dioxygen = Mole fraction $\times$ Total pressure
= $\frac{4}{5}\times 1=0.8\text{bar}$
Now, 1 bar = ${10}^{5}N{m}^{-2}$
So, 0.8 bar = $0.8\times {10}^{5}N{m}^{-2}$
Thus, the partial pressure of dioxygen would be $8\times {10}^{4}N{m}^{-2}$.

Theory:

Dalton's Law of partial pressure:

The total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of the individual gases.

Partial Pressure: It is the pressure exerted by a constituent gas of the gaseous mixture when kept alone in the same container.

Consider there are three non-reacting constituting gases 1, 2 and 3 whose partial pressures are $P_1,P_{2}and{P}_3$ respectively at temperature T and volume V. Then, mathematically, Dalton's Law of partial pressure can be written:

$P_1=\frac{\left(n_{1}RT\right)}{V}$

$P_2=\frac{\left(n_{2}RT\right)}{V}$

$P_3=\frac{\left(n_{3}RT\right)}{V}$

$P_T=P_1+P_2+P_3$

$P_T=\frac{(n_1+n_2+n_3)\left(RT\right)}{V}$

$n_1+n_2+n_3=n_t=\text{total number of moles}$

$P_T=n_t\left(\frac{RT}{V}\right)$

$P_T=P_1+P_2+P_3$ (at constant T, V)

where, $P_{Total}=\text{Total pressure exerted by the mixture of gases}$.

This law is only applicable for mixture of non-reacting gases

Note : It is also applicable for the reacting gases which are in chemical equilibrium.

By diving $\frac{P_1}{P_T},\frac{P_2}{P_T}and\frac{P_1}{P_T}$

We get :$\frac{P_1}{P_T}=n_1/n_T={\chi }_1$ (mole fraction of 1st gas)

$\frac{P_2}{P_T}=n_2/n_T={\chi }_2$(mole fraction of 2nd gas)

$\frac{P_3}{P_T}=n_3/n_T={\chi }_3$(mole fraction of 3rd gas)