Entropy of the Einstein Solid

1. Entropy of the Einstein Solid We consider the high-temperature case of q≫N. Consider now two Einstein solids with NA=300, NB=200, and qtotal=100 Aside from fluctuations that are normally much too small to measure, any isolated macroscopic system will inevitably evolve toward whatever (accessible) macrostate has the largest entropy. Equilibrium condition: http://www.compadre.org/stp/items/detail.cfm?ID=8685

2. Now let us ask how entropy is related to temperature. The most fundamental way to define temperature is in terms of energy flow and thermal equilibrium: Two objects in thermal contact are said to be at the same temperature if they are in thermal equilibrium, that is, if there is no spontaneous net flow of energy between them. If energy does flow spontaneously from one to the other, then we say the one that loses energy has the higher temperature while the one that gains energy has the lower temperature. So the quantity that is the same for both solids when they are in equilibrium is the slope dS/dq. Temperature must be some function of this quantity. To identify the precise relation between temperature and the slope of the entropy vs. energy, take qA that is larger than its equilibrium value. Here the entropy graph for solid B is steeper than that for solid A, meaning that if a bit of energy were to pass from A to B, solid B would gain more entropy than solid A loses. Since the total entropy would increase, the second law tells us that this process will happen spontaneously. In general, the steeper an object’s entropy vs. energy graph, the more it “wants” to gain energy (in order to obey the second law), while the shallower an object’s entropy vs. energy graph, the less it “minds” losing a bit of energy. We therefore conclude that temperature is inversely related to the slope dS/dU. In fact, the reciprocal (dS/dU)−1 has precisely the units of temperature, so we might guess simply In fact, this can be used as a definition of temperature: The factor of Boltzmann’s constant k in the definition of entropy eliminates the need for any further constants in this formula. To confirm that this relation gives temperature in ordinary Kelvin units one must check a particular example, as we will do in the following.

3. We consider the high-temperature Einstein solid (q≫N). consistent with the equipartition theorem Now we take the monoatomic ideal gas. If no work is performed (e.g., V=const) This equation works even if T=const, e.g., during the phase transition. When T is changing, we can write: What about S at T=0? Clausius 1865

4. Third law of thermodynamics (Nernst1906-1912) The third law of thermodynamics is sometimes stated as follows: The entropy of a perfect crystal at absolute zero is exactly equal to zero. At zero kelvin the system must be in a state with the minimum possible energy, and this statement of the third law holds true if the perfect crystal has only one minimum energy state. Entropy is related to the number of possible microstates, and with only one microstate available at zero kelvin, the entropy is exactly zero. A more general form of the third law applies to systems such as glasses that may have more than one minimum energy state (mathematically, the absolute entropy of any system at zero temperature is the natural log of the number of ground states times Boltzmann's constant): The entropy of a system approaches a constant value as the temperature approaches zero. The constant value (not necessarily zero) is called the residual entropy of the system. An example of a system which does not have a unique ground state is one containing half-integer spins, for which time-reversal symmetry gives two degenerate ground states. For such systems, the entropy at zero temperature is at least ln(2)k. Physically, the law implies that it is impossible for any procedure to bring a system to the absolute zero of temperature in a finite number of steps. The temperature of a substance can be reduced in an isentropic process by changing the parameter X from X2 to X1. Left: Absolute zero can be reached in a finite number of steps if S(0,X1)≠S(0, X2). Right: An infinite number of steps is needed since S(0,X1)= S(0,X2).

5. the heat capacity of all substances must go to zero at absolute zero (otherwise the integral would diverge; S is always non-negative)