assignment Homework

Energy, Entropy, and Probabilities

The goal of this problem is to show that once we have maximized the entropy and found the microstate probabilities in terms of a Lagrange multiplier \(\beta\), we can prove that \(\beta=\frac1{kT}\) based on the statistical definitions of energy and entropy and the thermodynamic definition of temperature embodied in the thermodynamic identity.

The internal energy and entropy are each defined as a weighted average over microstates: \begin{align} U &= \sum_i E_i P_i & S &= -k_B\sum_i P_i \ln P_i \end{align}: We saw in clase that the probability of each microstate can be given in terms of a Lagrange multiplier \(\beta\) as \begin{align} P_i &= \frac{e^{-\beta E_i}}{Z} & Z &= \sum_i e^{-\beta E_i} \end{align} Put these probabilities into the above weighted averages in order to relate \(U\) and \(S\) to \(\beta\). Then make use of the thermodynamic identity \begin{align} dU = TdS - pdV \end{align} to show that \(\beta = \frac1{kT}\).

assignment Homework

Energy, Entropy, and Probabilities

The goal of this problem is to show that once we have maximized the entropy and found the microstate probabilities in terms of a Lagrange multiplier \(\beta\), we can prove that \(\beta=\frac1{kT}\) based on the statistical definitions of energy and entropy and the thermodynamic definition of temperature embodied in the thermodynamic identity.

The internal energy and entropy are each defined as a weighted average over microstates: \begin{align} U &= \sum_i E_i P_i & S &= -k_B\sum_i P_i \ln P_i \end{align} We saw in clase that the probability of each microstate can be given in terms of a Lagrange multiplier \(\beta\) as \begin{align} P_i &= \frac{e^{-\beta E_i}}{Z} & Z &= \sum_i e^{-\beta E_i} \end{align} Put these probabilities into the above weighted averages in order to relate \(U\) and \(S\) to \(\beta\). Then make use of the thermodynamic identity \begin{align} dU = TdS - pdV \end{align} to show that \(\beta = \frac1{kT}\).

assignment Homework

Entropy of mixing

Suppose that a system of \(N\) atoms of type \(A\) is placed in diffusive contact with a system of \(N\) atoms of type \(B\) at the same temperature and volume.

1. Show that after diffusive equilibrium is reached the total entropy is increased by \(2Nk\ln 2\). The entropy increase \(2Nk\ln 2\) is known as the entropy of mixing.

2. If the atoms are identical (\(A=B\)), show that there is no increase in entropy when diffusive contact is established. The difference has been called the Gibbs paradox.

3. Since the Helmholtz free energy is lower for the mixed \(AB\) than for the separated \(A\) and \(B\), it should be possible to extract work from the mixing process. Construct a process that could extract work as the two gasses are mixed at fixed temperature. You will probably need to use walls that are permeable to one gas but not the other.

Note

This course has not yet covered work, but it was covered in Energy and Entropy, so you may need to stretch your memory to finish part (c).

assignment Homework

Free energy of a two state system

1. Find an expression for the free energy as a function of \(T\) of a system with two states, one at energy 0 and one at energy \(\varepsilon\).

2. From the free energy, find expressions for the internal energy \(U\) and entropy \(S\) of the system.

3. Plot the entropy versus \(T\). Explain its asymptotic behavior as the temperature becomes high.

4. Plot the \(S(T)\) versus \(U(T)\). Explain the maximum value of the energy \(U\).