# How to derive the Gibbs free energy for an ideal gas?

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

So I need to determine the fundamental relation for an ideal gas in the Gibbs free energy representation.

My thoughts:

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.

• 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 – getafix Nov 17 '16 at 4:41
• 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 – MathUser Nov 17 '16 at 5:12
• Okay. I think I get what you want, let me type out an answer for you :) – getafix Nov 17 '16 at 5:14
• @getafix This is the first time I have studied these subjects, sorry for my mistake – MathUser Nov 17 '16 at 5:14
• 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. – getafix Nov 17 '16 at 5:28

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 $S_0$

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.

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,$$G(T,P)=H(T_0,P_0)+C_p(T-T_0)-TS(T_0,P_0)-TC_p\ln{\frac{T}{T_0}}+TR\ln{\frac{P}{P_0}}$$