Free energy of an ideal gas

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 ($1~\mathrm{atm}$, $1~\mathrm{mol}$, $298~\mathrm{K}$, ...)




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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