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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.
At an isobaric process, entropy increases with temperature, as you provide heat to the system.
At an isothermic process, entropy decreases with pressure, as you provide work to the system, which is converted to heat, leaving the system.
At an isochoric process, entropy increases with temperature (which increases the pressure), as you provide heat to the 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 $C_p-C_v$.
