For an ideal gas of two-component ($A+B$) mixture at a temperature $T$, which expression is correct?

$$p_T=p_A+p_B \tag{1}$$ or $$\rho_T=\rho_A+\rho_B, \tag{2}$$

where $p_T$ and $\rho_T$ are the total pressure and total mass density, respectively. They are useful when defining the molar fraction $x_i$ and mass fraction $\omega_i$ of gas mixture:

$$x_i=\frac{p_i}{p_T}, \quad \omega_i=\frac{\rho_i}{\rho_T} \quad i=A \, \text{or} \, B.$$

If no chemical reaction between the two components, let's assume Eq.(1) is correct firstly. According to the ideal gas law, we have $p_A=\rho_A \frac{R}{M_A}T$ and $p_B=\rho_B \frac{R}{M_B}T$, where $M_A$ and $M_B$ are the molar mass of the two species, $R$ is the universal gas constant. Similarly, for the gas mixture, we have

$$p_T=\rho_T \frac{R}{M_T} T=\frac{\rho_T R T}{x_A M_A+(1-x_A)M_B},$$

where $x_A=p_A/p_T$ is the molar fraction of component $A$ and $M_T$ is the molar mass of the mixture. Substituting $p_A$, $p_B$ and $p_T$ into Eq.(1) and solve for $\rho_T$, we found that

$$\rho_T= (\frac{\rho_A}{M_A}+\frac{\rho_B}{M_B})[x_A M_A+(1-x_A)M_B]. \tag{3}$$

Equation (3) seems to be not consistent with Eq.(2).

I believe Eq.(1) should be correct, which states that the total pressure of an ideal gas mixture is the sum of the partial pressures of the gases in the mixture. I am confusing when I read the prestigious textbook Transport Phenomena by Bird, Stewart & Lightfoot. Here is a screenshot of the related page. enter image description here

I don't know what I am doing wrong in deriving Eq.(3). Can anyone please correct me? Thank you very much!


2 Answers 2


The formula 2 is not right. You proved it yourself. But you don't need a mathematical proof. Just think about it. If you add a little bit of heavy gas to one tank of hydrogen you get the same density as when you add equal volumes of heavy gas and hydrogen? Density is an example of intensive parameters. Temperature, resistivity, dielectric constant are other examples. The resultant property of the system is not the sum of the properties of the components. It depends on the specific combination of components. Other parameters do add up, like mass, or force and pressure.

  • $\begingroup$ Thank you! Please see my update. I understand your explanation but still feel confused with how we defined mass fraction with $\omega_i=\rho_i/\sum_i\rho_i$? Specifically, what is $\rho_i$ exactly? $\endgroup$
    – user55777
    Commented Dec 2, 2020 at 6:00
  • $\begingroup$ @user55777 isn’t it just $m_{i}/V$ where $V$ is the total volume of the mixture? $\endgroup$
    – Bob D
    Commented Dec 2, 2020 at 11:11

I know this answer is late, but what Bird, Stewart, and Lightfoot said is that the sum of the mass concentrations of each individual species in the mixture is equal to the mass density of the solution. It is important to understand that in mass transfer, mass concentration of a species 'i' is defined as the mass of species 'i' in the mixture per unit volume of the mixture (not of the species itself).

Furthermore, it is clear that the mass density of a solution must be the total mass of the solution divided by the volume of the mixture. Therefore, when summing the densities of each species, you are summing fractions with a common denominator (volume of the mixture, due to the definition of mass concentration), where the numerator of each is the mass of each species, adding up to the total mass on the numerator. You can prove this by basic mathematics. I hope this is clear.

Equation (2) is correct, then, as long as the definition of 𝜌i is defined as mass of species 'i' per unit volume of mixture. If you define it as mass of species 'i' per unit volume occupied by species 'i', then it is not correct, but that is not the definition used for mass transfer theory (I have not seen any textbook using the latter definition). As for why your proof was incorrect, I am not really sure actually.


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