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Figure 3.3 shows graphs of entropy vs. energy for two objects, A and B. Both graphs are on the same scale. The energies of these two objects initially have the values indicated; the objects are then brought into thermal contact with each other. Explain what happens subsequently and why, without using the word "temperature."

Short Answer

Expert verified

Since the slope ∂S∂Uis different for both objects, the combined system will evolve by exchanging energy until the slopes become equal.

Step by step solution

01

Concept Introduction

If two objects are in thermal equilibrium, their temperatures are said to be the same. It can be expressed in terms of the system's entropy and energy, the measure of randomness being the entropy. It is determined by the amount of energy that is not available for work.

02

Explanation

Mathematically, temperature can be defined as:

1T=∂S∂U

Where,

∂Sis the change in entropy and ∂Uis the change in the internal energy of the body.

Hence, a graph reflecting these values can be used to study the temperature of that body.

Conversely, it can also be said that if the slope ∂S∂Uof bodies are same, their temperatures would be same.

From the figure, it can be seen that the slope ∂S∂Uof A is greater than that of B. The system will evolve through energy exchange until the slopes are equal. Assume that UA+UB=Utotal, i.e. that energy is exchanged in both directions. As a result, if UAis increased, UBmust be reduced in equal measure, and vice versa. In this situation, one option to make the slope equal is to decrease the slope for B and increase the slope for A, which may be accomplished by reducing UBand raising by the same amount. System B will transfer some energy to A until the slopes are equal. The temperature is referred to as the slope in this case.

03

Final answer

As a result, the slope ∂S∂Uof A is greater than that of B, and the system will evolve by exchanging energy until the slopes are equal.

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Most popular questions from this chapter

In Problem 2.18 you showed that the multiplicity of an Einstein solid containing N oscillators and q energy units is approximately

Ω(N,q)≈q+Nqqq+NNN

(a) Starting with this formula, find an expression for the entropy of an Einstein solid as a function of N and q. Explain why the factors omitted from the formula have no effect on the entropy, when N and q are large.

(b) Use the result of part (a) to calculate the temperature of an Einstein solid as a function of its energy. (The energy is U=qϵ, where ϵis a constant.) Be sure to simplify your result as much as possible.

(c) Invert the relation you found in part (b) to find the energy as a function of temperature, then differentiate to find a formula for the heat capacity.

(d) Show that, in the limit T→∞, the heat capacity is C=Nk. (Hint: When x is very small, ex≈1+x.) Is this the result you would expect? Explain.

(e) Make a graph (possibly using a computer) of the result of part (c). To avoid awkward numerical factors, plot C/Nkvs. the dimensionless variable t=kT/ϵ, for t in the range from 0 to about 2. Discuss your prediction for the heat capacity at low temperature, comparing to the data for lead, aluminum, and diamond shown in Figure 1.14. Estimate the value of ϵ, in electron-volts, for each of those real solids.

(f) Derive a more accurate approximation for the heat capacity at high temperatures, by keeping terms through x3 in the expansions of the exponentials and then carefully expanding the denominator and multiplying everything out. Throw away terms that will be smaller than(ϵ/kT)2 in the final answer. When the smoke clears, you should find C=Nk1-112(ϵ/kT)2.

In Problem 1.55 you used the virial theorem to estimate the heat capacity of a star. Starting with that result, calculate the entropy of a star, first in terms of its average temperature and then in terms of its total energy. Sketch the entropy as a function of energy, and comment on the shape of the graph.

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?

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=4.77×10-3J/K2, and c=8.54×105J·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).

Polymers, like rubber, are made of very long molecules, usually tangled up in a configuration that has lots of entropy. As a very crude model of a rubber band, consider a chain of N links, each of length â„“(see Figure 3.17). Imagine that each link has only two possible states, pointing either left or right. The total length L of the rubber band is the net displacement from the beginning of the first link to the end of the last link.

(a) Find an expression for the entropy of this system in terms of N and NR, the number of links pointing to the right.
(b) Write down a formula for L in terms of N and NR.
(c) For a one-dimensional system such as this, the length L is analogous to the volume V of a three-dimensional system. Similarly, the pressure P is replaced by the tension force F. Taking F to be positive when the rubber band is pulling inward, write down and explain the appropriate thermodynamic identity for this system.
(d) Using the thermodynamic identity, you can now express the tension force F in terms of a partial derivative of the entropy. From this expression, compute the tension in terms of L, T, N, and â„“.
(e) Show that when L << Nâ„“, the tension force is directly proportional to L (Hooke's law).
(f) Discuss the dependence of the tension force on temperature. If you increase the temperature of a rubber band, does it tend to expand or contract? Does this behavior make sense?
(g) Suppose that you hold a relaxed rubber band in both hands and suddenly stretch it. Would you expect its temperature to increase or decrease? Explain. Test your prediction with a real rubber band (preferably a fairly heavy one with lots of stretch), using your lips or forehead as a thermometer. (Hint: The entropy you computed in part (a) is not the total entropy of the rubber band. There is additional entropy associated with the vibrational energy of the molecules; this entropy depends on U but is approximately independent of L.)

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