# Variation of entropy with volume under constant pressure

For a thermodynamic system defined with E and V (internal energy and volume), how would the entropy vary with the volume, under constant pressure?

I've managed to get dS = Cp * ln ( T2 / T1) , where Cp is the constant pressure specific heat capacity.

But I'm not sure how to relate volume to the equation. I'm guessing if the system if defined by only E and V, it would be alright to use ideal gas

Edit : I am wondering if I can make my way from the first law (d U=T d S-p d V+\mu d N ), and if it would be valid to assume that U does not change..

For a reversible process with an ideal gas, the molar entropy change is determined in one of two ways.

$$\Delta\bar{S}^\star = \int \bar{C}_V\ d\ln(T) + R \ln\left( \frac{V_f}{V_i}\right)$$

$$\Delta\bar{S}^\star = \int \bar{C}_p\ d\ln(T) - R \ln\left( \frac{p_f}{p_i}\right)$$

These formulations are derived in textbooks on thermodynamics. The specific heat capacity is not required to be constant.

For anything other than a non-ideal gas, the molar entropy change of a reversible process includes a term for non-ideality.

$$\Delta\bar{S} = \Delta\bar{S}^\star + \Delta\bar{S}_{non-ideal}$$

The non-ideal term can be stated as an EXCESS function or DEPARTURE function. Either approach is valid and both are documented in textbooks of thermodynamics (typically in advanced engineering thermodynamics). The non-ideality can also be expressed through $$p,\bar{V},T$$ and partial derivatives thereof. Ultimately then, one should be able to derive an expression for the non-ideality when given the mechanical $$(p,\bar{V},T)$$ equation of state for the substance. By example, for a compressibility model of a gas with $$Z \equiv p\bar{V}/RT$$, the following applies to determine the departure function of entropy of a real gas from an ideal gas:

$$\bar{S}(T,p) - \bar{S}^\star(T,p) = R \ln Z + \int_\infty^{\bar{V}} \left[\left(\frac{\partial p}{\partial{T}}\right)_{\bar{V}} - \left( \frac{R}{\bar{V}}\right) \right] d\bar{V}$$

A simple exercise is to prove this is identically zero for an ideal gas (as expected).

The expressions above are for reversible processes. Irreversible processes add their own $$\Delta\bar{S}_{irr}$$. This term is path dependent.

If it's an ideal gas, you can relate volume to temperature by the ideal gas law. That should be enough for you to proceed.

With regard to your edit, if it is an ideal gas your assumption that $$\Delta U=0$$ would not be valid, since $$\Delta U$$ depends only on temperature change and is $$\Delta U=C_{v}\Delta T$$. For $$\Delta U$$ to be zero for an ideal gas, $$T_{2}=T_{1}$$ and that would make $$\Delta S=0$$ in your equation.