Energy and Entropy: Fall-2020
HW 6: Due Friday 11/13

  1. Hot Metal Cube

    A 1 cm3 cube of hot metal is thrown into the ocean; several hours pass.

    1. During this time does the entropy of the metal increase, decrease, remain the same, or is this not determinable with the given information? Explain your reasoning.

    2. Does the entropy of ocean increase, decrease, remain the same, or is this not determinable with the given information? Explain your reasoning.

    3. Does the entropy of the metal plus the ocean increase, decrease, remain the same, or is this not determinable with the given information? Explain your reasoning.

  2. Bungee

    A physics major carefully measures the tension in a Bungee cord over a range of temperatures from room temperature to the boiling point of water. She examines her data carefully and finds that the tension in the cord is very well approximated by \begin{align*} \tau = \left(a - be^{-T/T_1}\right) \tan\left(\frac{\pi L}{2L_M}\right) \end{align*} where \(L\) is the length that the cord is stretched beyond its relaxed length, and \(a\), \(b\), \(T_1\) and \(L_M\) are positive constants.

    She then places the relaxed cord (\(L=0\)) in a calorimeter and measures the heat capacity over the same range of temperatures and finds that \begin{align*} C_L = \alpha + \gamma e^{-T/T_1} \end{align*} where \(\alpha\) and \(\gamma\) are two additional positive constants, and \(T_1\) is the same value found in the previous experiment.

    In this problem, you will need the thermodynamic identity for a one-dimensional system under tension: \begin{align} dU &= TdS+\tau dL \end{align} where \(\tau\) is the tension and \(L\) is the length. The Helmholtz free energy for a one-dimensional system is defined in the usual way as \(F=U-TS\).

    1. Sketch the tension \(\tau\) versus the stretch \(L\), and the heat capacity \(C_L\) versus the temperature \(T\).

    2. Find the change in Helmholtz free energy \begin{align*} \Delta F &= F(T, L) - F(T, \tfrac12 L_M) \end{align*}

    3. Solve for the change in entropy \(S(T,L) - S(T, \frac12 L_M)\) at an arbitrary temperature and length.

    4. Solve for the change in internal energy \(U(T,L) - U(T, \frac12 L_M)\) at an arbitrary temperature and length.

  3. Plastic Rod

    When stretched to a length L the tension force \(\tau\) in a plastic rod at temperature \(T\) is given by its Equation of State \begin{equation*} \tau = a T^{2} (L - L_{o}) \end{equation*} where \(a\) is a positive constant and \(L_{o}\) is the rod's unstretched length. For an unstretched rod (i.e. \(L = L_{o}\)) the heat capacity at constant length is \(C_{L}=bT\) where \(b\) is a constant. Knowing the internal energy at \(T_{o}, L_{o}\) (i.e. \(U(T_{o},L_{o})\)) find the internal energy \(U(T_{f},L_{f})\) at some other temperature \(T_{f}\) and length \(L_{f}\).

    1. Write \(U=U(T,L)\) and take the total differential \(dU\).

    2. Show that the partial derivative \((\partial U / \partial L)_{T} = -aT^{2}(L-L_{o})\).

    3. Integrate \(dU\) very carefully in the \(T, L\) plane, keeping in mind that \(C_{L} = bT\) holds only at \(L=L_{o}\) to find \(U(T_f,L_f)-U(T_0,L_0)\).

  4. Using Gibbs Free Energy

    You are given the following Gibbs free energy: \begin{equation*} G=-k T N \ln \left(\frac{a T^{5 / 2}}{p}\right) \end{equation*} where \(a\) is a constant (whose dimensions make the argument of the logarithm dimensionless).

    1. Compute the entropy.

    2. Work out the heat capacity at constant pressure \(C_p\).

    3. Find the connection among \(V\), \(p\), \(N\), and \(T\), which is called the equation of state (Hint: find the volume as a partial derivative of the Gibbs free energy).

    4. Compute the internal energy \(U\).

  5. Non-Ideal Gas

    The equation of state of a gas that departs from ideality can be approximated by \[ p=\frac{NkT}{V}\left(1+\frac{NB_{2}(T)}{V}\right) \] where \(B_{2}(T)\) is called the second virial coefficient which increases monotonically with temperature. Find \(\left( \frac{\partial U}{\partial V}\right)_{T}\) and determine its sign.