Gibbs free energy for an ideal gas (problem)

Question

so I'm trying to find the gibbs free energy for an ideal gas

$$g = u -Ts + pv $$

Hence,

$$dg = du -Tds -sdT +pdv +vdp = -sdt + vdp$$

The entropy is $$s = c_{p} \ln T -R \ln P + s_{o}$$

Which leads to

$$dg = \int{ dt (-c_{p} \ln T + R \ln P -s_{po})} + \int{vdp}$$

$$ = c_{p}T - c_{p}T \ln T + RT \ln P -s_{o}T + RT \ln P$$

Would anyone know why is there an extra $RT \ln P$ at the end, where did I go wrong?

Answer 1

There is an easier way:

$$g = h - sT$$

$$ h = c_p(T-T_0)+h_0$$

$$s = c_p\ln\frac{T}{T_0} - R \ln\frac{p}{p_0} + s_0 $$

Then put it all together:

$$ \boxed{ g = c_p (T-T_0) - T c_p\ln\frac{T}{T_0} + RT \ln\frac{p}{p_0} + h_0 - s_0 T_0 }$$

Answer 2

To me you can not integrate functions by considering that P or/and T are constants since they depend from one each other.

Indeed, the only way to integrate $\int vdP$ to $RTlnP$ is by considering that your transformation is isothermal. But on the first integral this time that P is constant to get $RTlnP$ from $\int RlnP$.

So on the first integral the change of free enthalpy is carried through isobar process and on the second integral the change of free enthalpy is carried through isothermal process.I guess the error comes from there..