The mixing entropy formula derived in the previous problem actually applies to any ideal gas, and to some dense gases, liquids, and solids as well. For the denser systems, we have to assume that the two types of molecules are the same size and that molecules of different types interact with each other in the same way as molecules of the same type (same forces, etc.). Such a system is called an ideal mixture. Explain why, for an ideal mixture, the mixing entropy is given by

ΔSmixing=klnNNA

where Nis the total number of molecules and NAis the number of molecules of type A. Use Stirling's approximation to show that this expression is the same as the result of the previous problem when both Nand NAare large.

Short Answer

Expert verified

The chane in entropy mixing for an ideal mixture isΔSmixingNk[(1x)ln(1x)+xlnx]

Step by step solution

01

Step: 1  Equating entropy:

Let's say we start with a set of Nidentical molecules. This system's entropy is a number S0, which may or may not be straightforward to determine. Assume that we suddenly transform NAof these molecules to a new species at some point in the future (which has similar properties to the original species as mentioned).The amount of different ways we may select to locate these NAmolecules among the Nsites accessible will increase the entropy.

The entropy of mixing is:

ΔSmixing=klnNNA

The Coefficient binomial is

ΔSmixing=klnN!NA!NNA!

where,NA=(1-x)N

ΔSmixing=klnN!((1x)N)!(xN)!

02

Step: 2 Stirling's approximation:

By using Stirling's approximationn!2πnnnen,the factorial is

N!2πNNNeNxN!2πxN(xN)xNexN(1x)N!2π(1x)N((1x)N)(1x)Ne(x1)N

Substitutiong values,we get

ΔSmixingkln2πNNNeN2π(1x)N((1x)N)(1x)Ne(x1)N2πxN(xN)xNexNΔSmixingkln2πNNN2π(1x)N((1x)N)(1x)N2πxN(xN)xNΔSmixingklnNN2πNx(1x)((1x)N)(1x)N(xN)xN

03

Step: 3 Finding entropy mixing value:

Where,ln(ab)=ln(a)+ln(b);lnab=ln(a)ln(b)

we get,

ΔSmixingkNlnN12ln(2πNx(1x))((1x)N)ln((1x)N)xNln(xN)

Taking third term part

((1x)N)ln((1x)N)=((1x)N)(ln(1x)+ln(N))((1x)N)ln((1x)N)=Nln(1x)+xNln(1x)Nln(N)+xNln(N)

Taking fourth term part

xNln(xN)=xNlnxxNlnN

The entropy mixing by

ΔSmixingk12ln(2πNx(1x))(1x)Nln(1x)xNlnxΔSmixingNk[(1x)ln(1x)+xlnx]

The first component in the second line has been omitted since it is insignificant in comparison to the following two terms for bigN.

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Most popular questions from this chapter

Consider again the system of two large, identical Einstein solids treated in Problem 2.22.

(a) For the case N=1023, compute the entropy of this system (in terms of Boltzmann's constant), assuming that all of the microstates are allowed. (This is the system's entropy over long time scales.)

(b) Compute the entropy again, assuming that the system is in its most likely macro state. (This is the system's entropy over short time scales, except when there is a large and unlikely fluctuation away from the most likely macro state.)

(c) Is the issue of time scales really relevant to the entropy of this system?

(d) Suppose that, at a moment when the system is near its most likely macro state, you suddenly insert a partition between the solids so that they can no longer exchange energy. Now, even over long time scales, the entropy is given by your answer to part (b). Since this number is less than your answer to part (a), you have, in a sense, caused a violation of the second law of thermodynamics. Is this violation significant? Should we lose any sleep over it?

The natural logarithm function, ln, is defined so that elnx=xfor any positive numberx.
aSketch a graph of the natural logarithm function.
b Prove the identities
localid="1650331641178" lnab=lna+lnbandlocalid="1650331643409" lnab=blna
(c) Prove thatlocalid="1650331645612" ddxlnx=1x.
(d) Derive the useful approximation

localid="1650331649052" ln(1+x)x

which is valid when localid="1650331651790" |x|1. Use a calculator to check the accuracy of this approximation for localid="1650331654235" x=0.1and localid="1650331656447" x=0.01.

For either a monatomic ideal gas or a high-temperature Einstein solid, the entropy is given by times some logarithm. The logarithm is never large, so if all you want is an order-of-magnitude estimate, you can neglect it and just say . That is, the entropy in fundamental units is of the order of the number of particles in the system. This conclusion turns out to be true for most systems (with some important exceptions at low temperatures where the particles are behaving in an orderly way). So just for fun, make a very rough estimate of the entropy of each of the following: this book (a kilogram of carbon compounds); a moose of water ; the sun of ionized hydrogen .

Use a computer to reproduce the table and graph in Figure2.4: two Einstein solids, each containing three harmonic oscillators, with a total of six units of energy. Then modify the table and graph to show the case where one Einstein solid contains six harmonic oscillators and the other contains four harmonic oscillators (with the total number of energy units still equal to six). Assuming that all microstates are equally likely, what is the most probable macrostate, and what is its probability? What is the least probable macrostate, and what is its probability?

How many possible arrangements are there for a deck of 52playing cards? (For simplicity, consider only the order of the cards, not whether they are turned upside-down, etc.) Suppose you start w e in the process? Express your answer both as a pure number (neglecting the factor of k) and in SI units. Is this entropy significant compared to the entropy associated with arranging thermal energy among the molecules in the cards?

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