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

Question

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.

Answer 1

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 $S_0$

Answer 2

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.

Answer 3

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