# Relationship between the partial pressure $p_J = x_J p$ and the perfect gas law $pV = nRT$

I am currently studying Atkins' Physical Chemistry, 11th edition, by Peter Atkins, Julio de Paula, and James Keeler. Chapter 1A.2 Equations of state says the following:

When dealing with gaseous mixtures, it is often necessary to know the contribution that each component makes to the total pressure of the sample. The partial pressure, $$p_J$$, of a gas $$J$$ in a mixture (any gas, not just a perfect gas), is defined $$p_J = x_J p \tag{1A.6}$$ where $$x_J$$ is the mole fraction of the component $$J$$, the amount of $$J$$ expressed as a fraction of the total amount of molecules, $$n$$, in the sample: $$x_J = \dfrac{n_J}{n} \ \ \ n = n_A + n_B + \dots \tag{1A.7}$$ When no $$J$$ molecules are present, $$x_J = 0$$; when only $$J$$ molecules are present, $$x_J = 1$$. It follows from the definition of $$x_J$$ that, whatever the composition of the mixture, $$x_A + x_B + \dots = 1$$ and therefore that the sum of the partial pressures is equal to the total pressure: $$p_A + p_B + \dots = (x_A + x_B + \dots)p = p \tag{1A.8}$$ This relation is true for both real and perfect gases.

When all the gases are perfect, the partial pressure as defined in eqn 1A.6 is also the pressure that each gas would exert if it occupied the same container alone at the same temperature.

It is this last part that I am unsure about:

When all the gases are perfect, the partial pressure as defined in eqn 1A.6 is also the pressure that each gas would exert if it occupied the same container alone at the same temperature.

I don't understand why this is true. Furthermore, is this not in reference to the perfect gas law, $$pV = nRT$$? Then how does this relate to $$pV = nRT$$?

I would greatly appreciate it if people would please take the time to clarify this.

• Well, if $p_{total}V=n_{total}RT$, and $n_A=n_{total}\cdot x_A$, then what is $p_A$? Commented Jun 16, 2020 at 5:49
• @IvanNeretin So $p_A = x_A p_{\text{total}} \Rightarrow p_{\text{total}} = \dfrac{p_A}{x_A}$ by 1A.6, and so we have that $\dfrac{p_A}{x_A} V = n_{\text{total}} RT$? Commented Jun 18, 2020 at 9:30
• @IvanNeretin And what about the other part of the question? Why is it that, when all the gases are perfect, the partial pressure as defined in eqn 1A.6 is also the pressure that each gas would exert if it occupied the same container alone at the same temperature? Commented Jun 18, 2020 at 9:32
• Why is it that if $2a=4$ then $a=2$? That's how algebra works. Commented Jun 18, 2020 at 9:40
• I think that the qualitative point is that when you refer to a 'perfect gas' you can not distinguish the identity of your atoms. There are equal and each one of them contributes equally to the total pressure. The rest is algebra.
– PAEP
Commented Jun 19, 2020 at 19:46

If the gases in the mixture are ideal, you can apply the ideal gas equation for it $$p_A V = n_ART$$where $$p_A$$ and $$n_A$$ represents the partial pressure and number of moles of a component A. Also $$p_tV = nRT$$ where $$p_t$$ and $$n$$ represents the total pressure and number of moles of the gaseous mixture. Dividing these equations you will get $$\frac{p_A}{p_t}=x_A$$ This equation will only be true for ideal gas mixture.
$$p_J = p_\mathrm{tot}x_J \tag{1}$$
as is done in Atkins, then the ratio $$p_J/p_\mathrm{tot} = x_J$$ is a definition and can't be wrong. The confusion comes from two different ways of defining partial pressure. Dalton's law uses a definition based on ideal gases, whereas the definition in Eq. 1 is based on actual parameters of the system. See this question for more on the topic.