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I want to calculate the Gibbs free energy for an ideal monoatomic gas, i.e. $$U=\frac32nRT.$$ Free energy is defined as $G(T,p)=H-TS$ and using the definition of enthalpy $$G(T,p)=U+pV-TS=U+nRT-TS.$$

To calculate the entropy I use the total differential of entropy given by $$\mathrm{d}S=\left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T +\left(\frac{\partial S}{\partial p}\right)_T \mathrm{d}p.$$

Using the Maxwell relation $$\left(\frac{\partial S}{\partial p}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_p,$$ and for an ideal gas with $V=\frac{nRT}{p}$ we hence get $$\left(\frac{\partial S}{\partial p}\right)_T=-\frac{nR}{p}.$$ Further $$\left(\frac{\partial S}{\partial T}\right)_p = \frac{c_p}{T}=\frac{\frac52nR}{T},$$ so all in all $$\mathrm{d}S=\frac{\frac52nR}{T}\mathrm{d}T-\frac{nR}{p}\mathrm{d}p.$$

How do I calculate $S$ from here, because integrating would leave me with $\Delta S$ which I cannot use?

The point of reference should be standard conditions ($1~\mathrm{atm}$, $1~\mathrm{mol}$, $298~\mathrm{K}$, ...)

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  • $\begingroup$ Why can't you use $\Delta S$? $\endgroup$ – Ivan Neretin Jan 17 '16 at 14:49
  • $\begingroup$ Well, because $G(T,p)=U+nRT-TS$ and I cannot just plug in the change in entropy there... $\endgroup$ – sj134 Jan 17 '16 at 14:51
  • $\begingroup$ Why do you need to have an absolute value of G. Isn't it sufficient to have G relative to some reference state? $\endgroup$ – Chet Miller Jan 18 '16 at 14:01
  • $\begingroup$ Are there no questions or comments about the answer I provided to this question? I obtained the result in two different ways, and got the same answer both ways. Do people feel that it is not correct? $\endgroup$ – Chet Miller Jan 19 '16 at 22:42
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Let $H^0$ be the heat of formation of the gas at $298~\mathrm{K}$ and $1\ \mathrm{atm}$. and let $G^0$ be the free energy of formation of the gas at these same conditions. The enthalpy of the gas at temperature $T$ (any pressure) is then: $$H(T)=H^0+C_p(T-298)$$ First let's get the free energy at temperature T and 1 atm. At constant pressure, the change in free energy with respect to temperature can be calculated from: $$\frac{d(G/T)}{dT}=-\frac{H}{T^2}$$ If we integrate this equation by parts, we obtain:

$$\frac{G(T,1~\mathrm{atm})}{T}-\frac{G^0}{298}=\frac{H(T)}{T}-\frac{H^0}{298}-\int_{298}^T{\frac{C_p}{T}dT}$$ Some algebraic manipulation of this equation yields: $$G(T,1~\mathrm{atm})=G^0+\frac{(G^0-H^0)}{298}(T-298)+C_p(T-298)-TC_p\ln(T/298)$$ If we define $S^0=\frac{(H^0-G_0)}{298}$, this equation becomes: $$G(T,1~\mathrm{atm})=G^0-S^0(T-298)+C_p(T-298)-TC_p\ln(T/298)$$ Adding the free energy change between $1~\mathrm{atm}$ and pressure $P$ at temperature $T$ yields: $$G(T,P)=G^0-S^0(T-298)+C_p(T-298)-T[C_p\ln(T/298)-R\ln(P/P_0)]\tag{1}$$ where $P_0=1~\mathrm{atm}.$ The term in brackets in this equation is the same as $S(T,P)-S^0$. If we substitute this into the previous equation, we obtain: $$G(T,P)=G^0-S^0(T-298)+C_p(T-298)-T[S(T,P)-S^0]$$ If we combine terms in this relationship, we obtain: $$G(T,P)-G^0=C_p(T-298)-(TS(T,P)-298S^0)$$ The term $C_p(T-298)$ is $\Delta H$, and the term $(TS(T,P)-298S^0)$ is $\Delta (TS)$. So, as expected, the equation is the same as $$G(T,P)-G^0=\Delta H-\Delta (TS)$$ where $$S=S^0+C_p\ln(T/298)-R\ln(P/P_0)]\tag{2}$$ Either Eqn. 1 or Eqn. 2 can be applied, depending on preference.

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