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.