Molar Concentration as a Function of Mole Fraction

Question

I am using the ideal gas law for a gas mixture and would like to confirm the following derivation and simplification.

Starting from the ideal gas law:

$$\rho = \frac{P \cdot M}{R \cdot T}$$

where:

• $\rho$ is the density of the gas mixture (in kg/m³);
• $P$ is the pressure (in Pa);
• $M$ is the mean molecular weight of the gas mixture (in kg/kmol);
• $R$ is the gas constant (in J/(kmol·K));
• $T$ is the temperature (in K).

Given the formula for the molar concentration $c_j$ of a component $\mathcal{S}_j$ of the gas mixture in terms of the density $\rho$ and the mole fraction $X_{\mathcal{S}_j}$:

$$ c_j = \frac{\rho \cdot X_{\mathcal{S}_j}}{M} $$

I substituted the expression for $\rho$ into the formula for $c_j$:

$$ c_j = \frac{\left(\frac{P \cdot M}{R \cdot T}\right) \cdot X_{\mathcal{S}_j}}{M} $$

Upon simplification, the $M$ terms cancel out:

$$ c_j = \frac{P \cdot X_{\mathcal{S}_j}}{R \cdot T} $$

My question is whether this simplification is correct. Does the mean molecular weight $M$ indeed cancel out in the expression for the molar concentration? Can the molar concentration $c_j$ be accurately expressed as:

$$ c_j = \frac{P \cdot X_{\mathcal{S}_j}}{R \cdot T} $$

without involving $M$?

Thank you for your help.

Answer 1

Your derivation is correct; however, that doesn't mean that you're eliminating the dependance of the molar concentration of species A on the average molar mass, since:

$$C=\frac{\rho}{M}=\frac{P}{RT}$$ $$C_\mathrm{A}=X_\mathrm{A}\;C=X_\mathrm{A}\frac{P}{RT}=X_\mathrm{A}\frac{\rho}{M}$$

Any change on $M$ will also change $C$ and $C_\mathrm{A}$.