Energy and Entropy: Fall-2020
HW 2: Due Friday 10/9

  1. Extensive Internal Energy

    Consider a system which has an internal energy \(U\) defined by: \begin{align} U &= \gamma V^\alpha S^\beta \end{align} where \(\alpha\), \(\beta\) and \(\gamma\) are constants. The internal energy is an extensive quantity. What constraint does this place on the values \(\alpha\) and \(\beta\) may have?

  2. Isolength and Isoforce Stretchability

    In class, you measured the isolength stretchability and the isoforce stretchability of your systems in the PDM. We found that for some systems these were very different, while for others they were identical.

    Show with algebra (NOT experiment) that the ratio of isolength stretchability to isoforce stretchability is the same for both the left-hand side of the system and the right-hand side of the system. i.e.: \begin{align} \frac{\left(\frac{\partial {x_L}}{\partial {F_L}}\right)_{x_R}}{\left(\frac{\partial {x_L}}{\partial {F_L}}\right)_{F_R}} &= \frac{\left(\frac{\partial {x_R}}{\partial {F_R}}\right)_{x_L}}{\left(\frac{\partial {x_R}}{\partial {F_R}}\right)_{F_L}} \label{eq:ratios} \end{align}

    Hint
    You will need to make use of the cyclic chain rule: \begin{align} \left(\frac{\partial {A}}{\partial {B}}\right)_{C} = -\left(\frac{\partial {A}}{\partial {C}}\right)_{B}\left(\frac{\partial {C}}{\partial {B}}\right)_{A} \end{align}
    Hint
    You will also need the ordinary chain rule: \begin{align} \left(\frac{\partial {A}}{\partial {B}}\right)_{D} = \left(\frac{\partial {A}}{\partial {C}}\right)_{D}\left(\frac{\partial {C}}{\partial {B}}\right)_{D} \end{align}

  3. Coffees and Bagels and Net Worth

    In economics, the term utility is roughly related to overall happiness. Many things affect your happiness, including the amount of money you have and the amount of coffee you drink. We cannot directly measure your happiness, but we can measure how much money you are willing to give up in order to obtain coffee or bagels. If we assume you choose wisely, we can thus determine that your happiness increases when you decrease your amount of money by that amount in exchange for increasing your coffee consumption. Thus money is a (poor) measure of happiness or utility.

    Money is also a nice quantity because it is conserved---just like energy! You may gain or lose money, but you always do so by a transaction. (There are some exceptions to the conservation of money, but they involve either the Fed, counterfeiters, or destruction of cash money, and we will ignore those issues.)

    In this problem, we will assume that you have bought all the coffee and bagels you want (and no more), so that your happiness has been maximized. Thus you are in equilibrium with the coffee shop. We will assume further that you remain in equilibrium with the coffee shop at all times, and that you can sell coffee and bagels back to the coffee shop at cost.*

    Thus your savings \(S\) can be considered to be a function of your bagels \(B\) and coffee \(C\). In this problem we will also discuss the prices \(P_B\) and \(P_C\), which you may not assume are independent of \(B\) and \(C\). It may help to imagine that you could possibly buy out the local supply of coffee, and have to import it at higher costs.

    1. The prices of bagels and coffee \(P_B\) and \(P_C\) have derivative relationships between your savings and the quantity of coffee and bagels that you have. What are the units of these prices? What is the mathematical definition of \(P_C\) and \(P_B\)?

    2. Write down the total differential of your savings, in terms of \(B\), \(C\), \(P_B\) and \(P_C\).

    3. Solve for the total differential of your net worth. Your net worth \(W\) is the sum of your total savings plus the value of the coffee and bagels that you own. From the total differential, relate your amount of coffee and bagels to partial derivatives of your net worth.

  4. Paramagnet (multiple solutions) We have the following equations of state for the total magnetization \(M\), and the entropy \(S\) of a paramagnetic system: \begin{align} M&=N\mu\, \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}} {e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}}\\ S&=Nk_B\left\{\ln 2 + \ln \left(e^{\frac{\mu B}{k_B T}}+e^{-\frac{\mu B}{k_B T}}\right) +\frac{\mu B}{k_B T} \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}} {e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}} \right\} \end{align}
    1. List variables in their proper positions in the middle columns of the charts below.

    2. Solve for the magnetic susceptibility, which is defined as: \[\chi_B=\left(\frac{\partial M}{\partial B}\right)_T \]

    3. Using both the differentials (zapping with d) and chain rule diagram methods, find a chain rule for:

      \[\left(\frac{\partial M}{\partial B}\right)_S \]

    4. Evaluate your chain rule. Sense-making: Why does this come out to zero?