Homework Assignment # 8

Due 22 Nov. 2019

Do the following problems: Chapter 8: 8-3, 8-4 [just derive the equation $\bar U(T,\bar V)- \bar U^{\rm id}(T)= -a\,/ \, \bar V$], 8-10 (do not do the part dealing with the Redlich-Kwong gas), 8-14, 8-41, 8-44

In addition, do the following problems:

  1. Eq. (8.22) relates $\partial U/\partial V$ at constant temperature to the three variables $P, V$, and $T$. Derive a similar equation for $\partial H/\partial P$ at constant T. Start with the definition $G=H-TS$. Then differentiate with respect to $P$ with $T$ constant: $$\left (\frac{\partial G}{\partial P}\right )_T = \left (\frac{\partial H}{\partial P}\right )_T - T \left (\frac {\partial S}{\partial P}\right )_T\,\,\,(1)$$ This can be rewritten as $$\left (\frac{\partial H}{\partial P}\right )_T=\left (\frac{\partial G}{\partial P}\right )_T+T \left (\frac {\partial S}{\partial P}\right )_T \,\,\,(2)$$ We know also that \begin{equation} \label{eq:dG} dG = - SdT + VdP\,\,\, (3) \end{equation} This comes from the 1st law $dU=TdS - PdV$, followed by the definition of $dU = dH +PdV + VdP$.
    Now, we use the mathematical expression for the exact differential of a function of two variables $$dG = \left (\frac {\partial G}{\partial T}\right )_P dT + \left (\frac {\partial G}{\partial P}\right )_T dP\,\,\,(4)$$ Equating Eqs. (3) and (4) gives $(\partial G/\partial T)_P = - S$ and $(\partial G/\partial P)_T = V$

    We can differential the first by $P$ holding $T$ constant, and the second by $T$, holding $P$ constant, and use the equivalence of the mixed second derivatives to obtain the Maxwell relation $$-\left (\frac{\partial S}{\partial P}\right )_T = \left (\frac{\partial V}{\partial P}\right )_T$$ Inserting this Maxwell relation into Eq. (2) leads to the desired result $$\left (\frac{\partial H}{\partial P}\right )_T= V - T\left (\frac{\partial V}{\partial T}\right )_P$$

  2. Some thermodynamic and physical properties of C(diamond) and C(graphite) are given in the following table
    quantity C(graphite) C(diamond)
    ΔHf(o) (kJ/mol) 0 1.897
    So (J/mol·K) 5.6 0.07
    CP (J/mol·K) 8.43 6.57
    ρ (g/cm3) 2.26 3.51

    Answer the following questions:

    1. What are the molar volumes (in m3/mol) of C(graphite) and C(diamond)? We're given here densities. If you invert the density, you get the volume taken up by one gram. Now, multiply this by the molar mass (12 g/mol for C) to get the volume taken up by one mole, which is the molar volume. We find $\bar V (graphite)$ = 5.310 cm3/mol and $\bar V (diamond)$ = 3.419 cm3/mol$

    2. For the reaction C(graphite) ⇋ C(diamond), at T=298 K, what is ΔGr(o).

      We know that G=H–TS so, at constant T, ΔG=ΔH– T ΔS So, ΔG=1.897e3–298×(0.07–5.6)=3.54 kJ/mol. At room temperature and 1 atm pressure, graphite is the stable allotrope of carbon.

    3. At T=298 K, at what pressure (in atm) will diamond become the more stable form of carbon? If we know $(\partial G/\partial P)_T$, we can calculate ΔG at any pressure P as $$G(P)=G(P=1\, {\rm atm})+\int_1^P (\partial G/\partial P)_T dP$$ We know that $(\partial G/\partial P)_T=V$, so $(\partial \Delta G/\partial P)_T=\Delta V$. Here, for one mole, ΔV = (3.419e-6 – 5.310e-6) m3/mol. Thus, \begin{eqnarray}\Delta G(P)&=&\Delta G(P=1\, {\rm atm})-\int_{1\,{\rm atm}}^P 1.891\times 10^{-6}dP \\ &=& 3.54 kJ/mol - 1.89\times 10^{-6} \int_1^P dP = 3.54 kJ/mol - 1.89\times 10^{-6} (P-1.01\times 10^5) \end{eqnarray} For diamond to be the stable form, we need $\Delta G(P) = 0$, Solving the equation 0=3.54e3– 1.89e-6(P–1.01e5) gives P=1.87e8 Pa = 1855 atm.

      Production of diamonds from graphite is facilitated by high pressure more than high temperature.