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When one is introduced to the first law of thermodynamics, one learns that enthalpy is only a function of temperature, and calculating its changes is straightforward \begin{equation} \Delta H^\pu{ig} = H^\pu{ig}(T_2) - H^\pu{ig}(T_1) = \int_{T_1}^{T_2} C_\mathrm{p}^\pu{ig}(T) \; dT \tag{1} \end{equation} When the second law is introduced, one also learns that the entropy is a function of pressure and temperature, and calculating its changes is also straightforward \begin{equation} \Delta S^\pu{ig} = S^\pu{ig}(p_2,T_2) - S^\pu{ig}(p_1,T_1) = \int_{T_1}^{T_2} \frac{C_\mathrm{p}^\pu{ig}(T)}{T} \; dT -R \ln\left(\frac{p_2}{p_1}\right) \tag{2} \end{equation} Finally, when the Gibbs energy is introduced, one is presented with the fundamental equation for constant composition systems $$ \mathrm{d}G = V\mathrm{d}p - S\mathrm{d}T \tag{3} $$ And Eq. (3) for an ideal gas, at constant temperature $T$, gives also a famous relationship to calculate the changes in pressure. We replace $V^\pu{ig} = RT/p$, integrate, and get \begin{equation} \Delta G^\pu{ig} = G^\pu{ig}(p_2,T) - G^\pu{ig}(p_1,T) = RT \ln\left(\frac{p_2}{p_1}\right) \tag{4} \end{equation}

Now, I am interested in calculating the changes in Gibbs energy when the temperature changes and the pressure is constant. However, one faces the funny situation of being unable to obtain an expression in this case. For a constant pressure process, integrating Eq. (3) is not particularly attractive since one gets the entropy in the right-hand side \begin{equation} \left(\frac{\partial G}{\partial T}\right)_p = -S \tag{5} \end{equation} It is possible to manipulate Eq. (5) and obtain the Gibbs-Helmholtz equation to eliminate $S$ for $H$ \begin{equation} \left[\frac{\partial (G^\pu{ig}/T)}{\partial T}\right]_p = -\frac{H^\pu{ig}}{T^2} \tag{6} \end{equation} and, for example, try to establish a relationship in a simpler case, of a constant heat capacity gas. Eq. (1) integrated for a reference temperature $T_0$ automatically leads to \begin{equation} \left[\frac{\partial (G^\pu{ig}/T)}{\partial T}\right]_p = -\frac{H^\pu{ig}(T_0) + C_\mathrm{p}^\pu{ig}(T - T_0)}{T^2} \tag{7} \end{equation} And no matter what the effort, I can never obtain a simple expression that only depends on the fundamental variables $p$ and $T$ $$ \Delta G^\pu{ig} = G^\pu{ig}(p,T_2) - G^\pu{ig}(p,T_2) = f(p,T_2) - f(p,T_1) $$ And thus, the changes of $\Delta G^\pu{ig}$ for a non-isothermal process in an ideal gas are impossible to calculate. A similar argument can be raised for calculating the changes in the Helmholtz energy $\Delta A^\pu{ig}$ when the temperature changes, and again, one cannot find an equation.

Is this correct? And if it is, why is so?

Edit Doing by definition of $\Delta G$ won't work because one faces an issue in the entropy term, impossible to resolve \begin{align} \Delta G^\pu{ig} &= [H^\pu{ig}(T_2) - T_2S^\pu{ig}(p_2,T_2)] - [H^\pu{ig}(T_1) - T_1S^\pu{ig}(p_1,T_1)] \\ &= [H^\pu{ig}(T_2) - H^\pu{ig}(T_1)] - \color{red}{[T_2S^\pu{ig}(p_2,T_2) - T_1S^\pu{ig}(p_1,T_1)]} \end{align}

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  • $\begingroup$ It seems $\int_{T_1=0}^{T_2} {\frac{C_p \mathrm{d}T}{T}}$ is easier for real matter, as $\lim_{T_1 \to 0} {(\frac{C_p}{T})} = 0$ for it, but $\lim_{T_1 \to 0} {(\frac{C_p}{T})} = \infty$ for ideal gas. $\endgroup$
    – Poutnik
    May 23, 2023 at 8:19
  • $\begingroup$ @Poutnik Hello Poutnik. In the first case you speak about solids, this is, the Debye model of solids that states $C_V \propto T^3$ and continuous at $T\rightarrow 0^+$? Is this right? And for the second case, how do you see that $T=0$ is making a problem in trying to find $\Delta G$? $\endgroup$ May 23, 2023 at 18:42
  • $\begingroup$ YEst, I do mean the Debye model. Integral dT/T from T=0 K is infinite, in attempts to get S as integration for T=0 K for ideal gas, as C_p,ig, does not go to zero. $\endgroup$
    – Poutnik
    May 23, 2023 at 19:08
  • $\begingroup$ @Poutnik So, if we don't include vibrational modes of motion, then $C_\mathrm{p}^\pu{ig} \neq f(T)$ and thus $\nexists \lim_{a \rightarrow 0^+} \int_a^{T_2} \frac{C_\mathrm{p}^\pu{ig}}{T} \; \mathrm{d}T$. You smell that this is the source of the problem? $\endgroup$ May 23, 2023 at 22:56
  • $\begingroup$ Imagine helium is an ideal gas, then it will still have Cp=3/2.R + R J/K/mol. / $\lim_{T \to 0+}{(\frac {dS}{dT})}=+\infty$ // I do not say it is the problem source, but rather additional problem. My TD is after leaving university 34 years ago quite dusty. $\endgroup$
    – Poutnik
    May 24, 2023 at 5:11

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