Suppose you have a mixture of gases (such as air, a mixture of nitrogen and oxygen). The mole fraction xiof any species iis defined as the fraction of all the molecules that belong to that species: xi=Ni/Ntotal. The partial pressure Piof species iis then defined as the corresponding fraction of the total pressure: Pi=xiP. Assuming that the mixture of gases is ideal, argue that the chemical potential μiof species i in this system is the same as if the other gases were not present, at a fixed partial pressure Pi.

Short Answer

Expert verified

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

01

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=NkTV

Here, Pis the pressure, Nis the number of species, kis the Boltzmann's constant, Tis the temperature and Vis the volume of the ideal gas.

02

Partial Pressure

Let us write the expression for the Partial pressure of ithspecies.

Pi=xiP

Here, Piis the Partial pressure of tthspecies and xiis the state of the ithspecies.

Substitute NkTVfor Pin the above expression.

Pi=xiNkTV

Substitute Nifor xiNin the above expression.

Pi=NikTV

Here, Niis the number of species in the xistate.

Consider a mixture of two ideal gases A and B.

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

St=SAUA,V,NA+SBUB,V,NB

03

Explanation

Here, Stis the total entropy of the mixture, SAand SBare the entropies of the gases A and B,NAand NBare the number of species in the gases A&Band Vis the volume.

The two entropies SAand SBare similar just like as only one gas is present in the mixture.

Write the expression for the chemical potential of gas A.

μA=-TSNAU,V.NB

Here, μAis 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.

μA=-TSNAU,V

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

04

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.

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Most popular questions from this chapter

A liter of air, initially at room temperature and atmospheric pressure, is heated at constant pressure until it doubles in volume. Calculate the increase in its entropy during this process.

Use the thermodynamic identity to derive the heat capacity formula

CV=TSTV

which is occasionally more convenient than the more familiar expression in terms of U. Then derive a similar formula for CP, by first writing dHin terms of dSand dP.

Estimate the change in the entropy of the universe due to heat escaping from your home on a cold winter day.

In Problem 2.32you computed the entropy of an ideal monatomic gas that lives in a two-dimensional universe. Take partial derivatives with respect to U,A, and N to determine the temperature, pressure, and chemical potential of this gas. (In two dimensions, pressure is defined as force per unit length.) Simplify your results as much as possible, and explain whether they make sense.

An ice cube (mass 30g)0°Cis left sitting on the kitchen table, where it gradually melts. The temperature in the kitchen is 25°C.

(a) Calculate the change in the entropy of the ice cube as it melts into water at 0°C. (Don't worry about the fact that the volume changes somewhat.)

(b) Calculate the change in the entropy of the water (from the melted ice) as its temperature rises from 0°Cto 25°C.

(c) Calculate the change in the entropy of the kitchen as it gives up heat to the melting ice/water.

(d) Calculate the net change in the entropy of the universe during this process. Is the net change positive, negative, or zero? Is this what you would expect?

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