Suppose I have a ideal gas mixture consisting of two different types of ideal gases, say $A$ and $B$, in a container with volume $V_{tot}$. There are $N_A$ particles of gas $A$ and $N_B$ particles of gas $B$. The temperature and pressure of the mixture are denotes by $T$ and $P$. Can I still use the ideal gas law for each of the component? Take $A$ as an example, is the following true?

$$P_A=\rho_A \frac{R}{M_A}T,$$ and $$V_{\text{m},A}=\frac{V_{tot}}{N_A}=\frac{RT}{P_A}=\frac{M_A}{\rho_A}$$

where $P_\text{A}$ is the partial pressure, $M_A$ is the molar mass, $V_{\text{m},A}$ is the partial molar volume, and $\rho_A$ is the partial density of $A$, i.e., the mass of species $A$ per unit volume of the mixture gas. $R$ is the universal gas constant. Thanks a lot!


Remember that ideal gases do not interact. They don't interact with themselves or with other ideal gases, so in a mixture of ideal gases each component of the mixture behaves as if the other gases were not present.

So for every type of gas present the ideal gas equation holds:

$$ P_i = n_i\frac{RT}{V} \tag{1} $$

And the total pressure is just the sum of all these partial pressures:

$$ P_{tot} = \sum_i P_i = \sum_i n_i\frac{RT}{V} \tag{2} $$

This can be rewritten in various ways using:

$$ \rho_i = \frac{M_i n_i}{V} $$

where $M_i$ is the molecular weight of the gas $i$, so $M_i n_i$ is the total mass present of species $i$. For example we can rewrite this as:

$$ \frac{V\rho_i}{M_i} = n_i $$

and susbtitute this into equation (1) to get the equation you give:

$$ P_i = n_i\frac{RT}{V} = \frac{V\rho_i}{M_i}\frac{RT}{V} = \frac{\rho_i}{M_i}RT $$

However it isn't useful to talk about a partial molar volume as all the gas molecules are uniformly distributed. It would only make sense to talk about $V_i$ if the two gases were separated by a movable partition so they were at the same temperature and pressure but had different volumes:


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  • $\begingroup$ @ John Rennie thanks for your rapid reply! But I want to involve molar mass of $A$, partial molar volume and partial density, $M_A$, $V_{m, A}$ and $\rho_A$ instead of those of the mixture gas. That is why I ask this question... Could you please comment the expression in my post? Thanks! $\endgroup$ – jsxs Apr 30 '17 at 7:55
  • $\begingroup$ @jsxs: is this better? $\endgroup$ – John Rennie Apr 30 '17 at 9:04
  • $\begingroup$ @ John Rennie, Thanks! it is helpful. But I still think it makes sense to talk about a molar volume of one component in a mixture gas, if we define it as the volume of this species per unit volume of the mixture gas, as seen in $V/n_i=M_i/\rho_i$ $\endgroup$ – jsxs Apr 30 '17 at 11:34

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