# What is the entropy change in isochoric process

I have studied that entropy increases with increase in temperature and it decreases with increase in pressure but in case of isochoric process both are happening at the same time but still the overall entropy of the system is somehow increased.

I want to know why the increase in entropy with increase in pressure dominates the decrease in entropy with increase in pressure since both are happening at the same time, resulting in an overall increase in entropy of system.

With implied constant heat capacity $$C_V$$:
$$\Delta S = \int_{T_1}^{T_2}{\mathrm{d}S}=\int_{T_1}^{T_2}{\frac{\delta Q}{T}}=\int_{T_1}^{T_2}{\frac{C_V\cdot \mathrm{d}T}{T}}=C_V \cdot \ln{\frac{T_2}{T_1}}$$
For an arbitrary differential change of a single phase substance, we have $$dS=C_p\frac{dT}{T}-\left(\frac{\partial V}{\partial T}\right)_PdP$$For an isochoric process, $$dV=\left(\frac{\partial V}{\partial T}\right)_PdT+\left(\frac{\partial V}{\partial P}\right)_TdP=0$$or$$dP=-\frac{\left(\frac{\partial V}{\partial T}\right)_P}{\left(\frac{\partial V}{\partial P}\right)_T}dT$$So, for an isochoric process, $$dS=C_p\frac{dT}{T}+\frac{[\left(\frac{\partial V}{\partial T}\right)_P]^2}{\left(\frac{\partial V}{\partial P}\right)_T}dT=\left[C_p-T\left(\frac{\partial V}{\partial T}\right)_P\left(\frac{\partial P}{\partial T}\right)_V\right]\frac{dT}{T}$$For all materials, the second term in brackets is always less than $$C_p$$, and is equal to $$C_p-C_v$$.