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Use the definition of temperature to prove the zeroth law of thermodynamics, which says that if system A is in thermal equilibrium with system B, and system B is in thermal equilibrium with system C, then system A is in thermal equilibrium with system C. (If this exercise seems totally pointless to you, you're in good company: Everyone considered this "law" to be completely obvious until 1931, when Ralph Fowler pointed out that it was an unstated assumption of classical thermodynamics.)

Short Answer

Expert verified

When the slopes of all three systems A, B, and C are equal, the systems are considered to be in thermal equilibrium, or at the same temperature.

Step by step solution

01

Concept Introduction

Temperature is a measure of hotness or coldness represented on one of several arbitrary scales that describes the natural flow of heat energy.

02

Explanation

If two items are in thermal equilibrium, their temperatures are said to be the same. The entropy and energy of a system can be used to express temperature. The gauge of randomness is entropy. The Zeroth law states that if a system A and a system B are in thermal equilibrium with another system C individually, then systems A and B will be in thermal equilibrium as well.

Mathematically, temperature can be stated as:

1T=∂S∂U

By rearranging the terms,

T=∂U∂S

The entropy vs energy graph will have equal slopes if the system is in thermal equilibrium. Because any two systems in thermal equilibrium have the same ∂S∂Uvalues, system B and C must have the same slope as system A, resulting in their slopes being identical. The zeroth law, which asserts that a system A can be set in thermal equilibrium with any other systems that are all in thermal equilibrium with each other, is the foundation of thermodynamics.

03

Final answer

As a result, if the slopes of all three systems A, B, and C are similar, the systems are said to be in thermal equilibrium, or at the same temperature.

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Most popular questions from this chapter

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).

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).

In order to take a nice warm bath, you mix 50 liters of hot water at 55°C with 25 liters of cold water at 10°C. How much new entropy have you created by mixing the water?

Starting with the result of Problem 2.17, find a formula for the temperature of an Einstein solid in the limit q≪N. Solve for the energy as a function of temperature to obtain U=Nϵe-ϵ/kT (whereϵ is the size of an energy unit).

Consider a monatomic ideal gas that lives at a height z above sea level, so each molecule has potential energy mgzin addition to its kinetic energy.

(a) Show that the chemical potential is the same as if the gas were at sea level, plus an additional term mgz:

μ(z)=-kTlnVN2πmkTh23/2+mgz.

(You can derive this result from either the definition μ=-T(∂S/∂N)U,Vor the formula μ=(∂U/∂N)S,V.

(b) Suppose you have two chunks of helium gas, one at sea level and one at height z, each having the same temperature and volume. Assuming that they are in diffusive equilibrium, show that the number of molecules in the higher chunk is

N(z)=N(0)e-mgz/kT

in agreement with the result of Problem 1.16.

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