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Determine the Gibbs Free Energy for an ideal gas and show that, this energy, and its derivate can be related to the enthalpy of the system via Gibbs-Helmholtz equation $$H = -T^2 \left( \frac{\partial (G/T)}{\partial T} \right)$$

I need determine the Gibbs free energy for an ideal gas. I know the Gibbs free energy, is defined by $$G(p,T) = U + pV - Ts = H - TS,$$ where $H$ is the enthalpy, then \begin{align} \mathrm{d}(U + pV - TS) &= V\mathrm{d}p - S\mathrm{d}T + \sum{\mu_i\mathrm{d}N_i} - \sum{X_i\mathrm{d}a_i}\\ \mathrm{d}G &= V\mathrm{d}p - S\mathrm{d}T + \sum{\mu_i\mathrm{d}N_i - \sum{X_i\mathrm{d}a_i}}\\ \therefore \mathrm{d}G &= V\mathrm{d}p - S\mathrm{d}T \end{align} Now, if the temperature is constant, and only the pressure change then $\mathrm{d}G = V\mathrm{d}p$. It follows that $$G(p)-G(p_0)=\int_{p_0}^{p}{V\mathrm{d}p}$$ If $V = \frac{nRT}{p}$, we have $$G(p) = G(p_0) + nRT \ln\left(\frac{p}{p_0}\right)$$

But, I don't know if this is correct.

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    $\begingroup$ I'm not quite sure what this question is asking for ? I don't know what the term "Gibbs Free Energy for an ideal gas" means. What the OP has derived here is the Gibbs Free Energy as a function of pressure, at constant temperature & mole number for an ideal gas. Initially my thoughts were, the OP desired the fundamental relation for an ideal gas in the gibbs free energy representation $\endgroup$ – getafix Nov 17 '16 at 4:41
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    $\begingroup$ Hi, thanks for your answer. The Gibbs free energy is defined as: $G(p.T)=U+pV-TS$. So, I need find the expression $G(p,T)$ for a ideal gas. The other part of the problem is not difficult. Thanks $\endgroup$ – MathUser Nov 17 '16 at 5:12
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    $\begingroup$ Okay. I think I get what you want, let me type out an answer for you :) $\endgroup$ – getafix Nov 17 '16 at 5:14
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    $\begingroup$ @getafix This is the first time I have studied these subjects, sorry for my mistake $\endgroup$ – MathUser Nov 17 '16 at 5:14
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    $\begingroup$ I wrote a quick answer, to your post. Please edit your question to include the clarification you made in the comments. Ask me if something is unclear, I have to go now. I will fill in some steps later perhaps. $\endgroup$ – getafix Nov 17 '16 at 5:28
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The fundamental relation for an Ideal Gas, in the entropy representation is

$$S(U,V,N) = NS_0 + NR\ln\left[ \left( \frac{U}{U_0} \right)^c \left( \frac{V}{V_0} \right) \left( \frac{N_0}{N} \right)^{c+1} \right]$$

The Gibbs Free Energy, is the Legendre Transform $G \equiv U[T,p]$

so, $G = U + pV -TS$

First, let's get the equation of state $S(T,V,N) $ from the the fundamental relation. We eliminate $U$ using $U = cNRT$

$$S(T,V,N) = NS_0 + NR\ln\left[ \left( \frac{T}{T_0} \right)^c \left( \frac{V}{V_0} \right) \left( \frac{N_0}{N} \right) \right].$$

Also, for an Ideal Gas, $p =\frac{NRT}{V}$

$$G(T,p,N) = cNRT - TNS_0 - NRT\ln\left[ \left( \frac{T}{T_0} \right)^c \left( \frac{p_0}{p} \right) \left( \frac{N_0}{N} \right) \right] + NRT $$

Exercise: Find $G_0 = G(T_0,p_0, N_0)$ and substitute into the fundamental relation derived above to get rid of $S_0$

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Yes you are partly on the right track. Your final equation is usually written in a more familiar form as $$G=G^{\ce{o} }+RT\ln\left(\frac{p}{p^{\ce{o}}}\right)$$ and letting $p^{\ce{o}}$ equal to $1$ atm then p is understood to be dimensionless and produces the free energy change with pressure $$G=G^{\ce{o} }+RT\ln(p)$$ However, this does not help, and to get to your result you quote the basic equation you need which is $dG=Vdp-SdT$ and at constant pressure $$\left (\frac {\partial G}{dT} \right)_p =-S $$ (The eqn you need is at constant p although it is not indicated in the equation in your question). Next, as $$G=H-TS$$ therefore $$G=H+T\left(\frac{\partial G}{dT}\right)_p$$

dividing through by $T^2$, working out the derivative $d(G/T)/dT$) and rearranging produces the Gibbs-Helmholtz equation.

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I think what they had in mind was for you to start with $$H(T,P)=H(T_0,P_0)+C_p(T-T_0)$$ and $$S(T,P)=S(T_0,P_0)+C_p\ln{\frac{T}{T_0}}-R\ln{\frac{P}{P_0}}$$So,$$G(T,P)=H(T_0,P_0)+C_p(T-T_0)-TS(T_0,P_0)-TC_p\ln{\frac{T}{T_0}}+TR\ln{\frac{P}{P_0}}$$

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