Question

Suppose you have a mixture of gases (such as air, a mixture of nitrogen and oxygen). The mole fraction $x_i$of any species $i$is defined as the fraction of all the molecules that belong to that species: $x_i=N_i/N_{\text{total}}$. The partial pressure $P_i$of species $i$is then defined as the corresponding fraction of the total pressure: $P_i=x_{i}P$. Assuming that the mixture of gases is ideal, argue that the chemical potential ${\mu }_i$of species $i$ in this system is the same as if the other gases were not present, at a fixed partial pressure $P_i$.

Short answer

The chemical potential for any number of species in the system of a mixture of ideal gases is equivalent to a system with only one gas at a constant partial pressure.

Step-by-step solution

Concept Introduction

The system of gases is an ideal gas so the total pressure of the system is equivalent to the partial pressure of any of its component present in the system. Also, the system will have no effect on the initially present components with the addition of more components while partial pressure is made constant. Therefore, a mixture of two ideal gases must be considered for such a case.

Let us write the expression for the Pressure of an ideal gas.

$P=\frac{NkT}{V}$

Here, $P$is the pressure, $N$is the number of species, $k$is the Boltzmann's constant, $T$is the temperature and $V$is the volume of the ideal gas.

Partial Pressure

Let us write the expression for the Partial pressure of $i^{\text{th}}$species.

$P_i=x_{i}P$

Here, $P_i$is the Partial pressure of $t^{\text{th}}$species and $x_i$is the state of the $i^{th}$species.

Substitute $\frac{NkT}{V}$for $P$in the above expression.

$P_i=x_i\frac{NkT}{V}$

Substitute $N_i$for $x_{i}N$in the above expression.

$P_i=\frac{N_{i}kT}{V}$

Here, $N_i$is the number of species in the $x_i$state.

Consider a mixture of two ideal gases A and B.

Write the expression for the total entropy of the mixture of the two gases.

$S_t=S_A\left(U_A,V,N_A\right)+S_B\left(U_B,V,N_B\right)$

Explanation

Here, $S_t$is the total entropy of the mixture, $S_A$and $S_B$are the entropies of the gases A and $B,N_A$and $N_B$are the number of species in the gases $A\&B$and $V$is the volume.

The two entropies $S_A$and $S_B$are similar just like as only one gas is present in the mixture.

Write the expression for the chemical potential of gas A.

${\mu }_A=-T{\left(\frac{\partial S}{\partial N_A}\right)}_{U,V.N_B}$

Here, ${\mu }_A$is the chemical potential of gas A and S is the entropy of the system.

Rewrite the above expression for the chemical potential of gas A as if gas B is not present in the mixture.

${\mu }_A=-T{\left(\frac{\partial S}{\partial N_A}\right)}_{U,V}$

The chemical potential of one component does not get affected by the other component in the mixture of two ideal gases.

Conclusion

Thus, the chemical potential for any number of species in the system of a mixture of ideal gases is equivalent to a system with only one gas at a constant partial pressure.