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Chapter 3: Q. 3.35 (page 119)

In the text I showed that for an Einstein solid with three oscillators and three units of energy, the chemical potential is 渭=-系(where 系is the size of an energy unit and we treat each oscillator as a "particle"). Suppose instead that the solid has three oscillators and four units of energy. How does the chemical potential then compare to -系 ? (Don't try to get an actual value for the chemical potential; just explain whether it is more or less than -系.)

Short Answer

Expert verified

渭<-鈭�

Step by step solution

01

Explanation of Solution

Given:

For N=3and q=3the chemical potential is 渭=-鈭�

02

Calculation

The chemical potential formula is

渭=螖U螖NS

The entropy for N=3and q=3is

S=kln惟

S=kln鈭�N+q-1|q-1|q鈭�3+3-1

S=kln|3-1|3鈭�5炉

S=kln|2|3|2||2|

S=kln(10)

The entropy must be constant throughout.

The entropy for N=3and q=4is,

S=kln鈦�惟S=kln鈦�鈭�N+q鈭�1|q鈭�1|q鈭�3+3鈭�1S=kln鈦�|3鈭�1|3鈭�5_S=kln鈦�|2|3|2||2|S=kln鈦�(10)

Thus, the entropy increases so to reduce the entropy to its original value.

03

Conclusion 

The chemical potential is lowered.

渭<-鈭�

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Most popular questions from this chapter

Use a computer to study the entropy, temperature, and heat capacity of an Einstein solid, as follows. Let the solid contain 50 oscillators (initially), and from 0 to 100 units of energy. Make a table, analogous to Table 3.2, in which each row represents a different value for the energy. Use separate columns for the energy, multiplicity, entropy, temperature, and heat capacity. To calculate the temperature, evaluate 螖U/螖Sfor two nearby rows in the table. (Recall that U=q系for some constant 系.) The heat capacity (螖U/螖T)can be computed in a similar way. The first few rows of the table should look something like this:

(In this table I have computed derivatives using a "centered-difference" approximation. For example, the temperature .28is computed as 2/(7.15-0).) Make a graph of entropy vs. energy and a graph of heat capacity vs. temperature. Then change the number of oscillators to 5000 (to "dilute" the system and look at lower temperatures), and again make a graph of heat capacity vs. temperature. Discuss your prediction for the heat capacity, and compare it to the data for lead, aluminum, and diamond shown in Figure 1.14. Estimate the numerical value of 蔚in electron-volts, for each of those real solids.

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

(a) Calculate the change in the entropy of the ice cube as it melts into water at 0掳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掳Cto 25掳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?

In solid carbon monoxide, each CO molecule has two possible orientations: CO or OC. Assuming that these orientations are completely random (not quite true but close), calculate the residual entropy of a mole of carbon monoxide.

Experimental measurements of heat capacities are often represented in reference works as empirical formulas. For graphite, a formula that works well over a fairly wide range of temperatures is (for one mole)

CP=a+bT-cT2

where a=16.86J/K,b=4.77脳10-3J/K2, and c=8.54脳105J路K. Suppose, then, that a mole of graphite is heated at constant pressure from 298Kto 500K. Calculate the increase in its entropy during this process. Add on the tabulated value of S(298K)(from the back of this book) to obtain S(500K).

As shown in Figure 1.14, the heat capacity of diamond near room temperature is approximately linear in T. Extrapolate this function up to 500K, and estimate the change in entropy of a mole of diamond as its temperature is raised from298K to 500K. Add on the tabulated value at298K (from the back of this book) to obtain S(500K).
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