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We know that internal energy and enthalpy of ideal gas is independent of pressure. The entropy of ideal gas is pressure dependent.

How about an in-compressible ($dv=0$) fluid? How can we use thermodynamic relations to investigate this?

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The internal energy of an incompressible liquid depends only on temperature. We know this from $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$Since dV=0, $dU=C_vdT$. From the form of the above equation, we can also show that Cv is independent of V (and a function only of T). So, $$\Delta U=\int{C_v(T')dT'}$$where T' is a dummy variable of integration. From the definition of enthalpy, we also have that: $$\Delta H=\Delta U+V\Delta P$$So, although internal energy of an incompressible fluid is dependent only on temperature, enthalpy of an incompressible fluid varies linearly with pressure. Another interesting feature is that, since the heat capacity at constant pressure is equal to the partial derivative of enthalpy with respect to temperature at constant pressure, the heat capacity at constant pressure of an incompressible fluid is equal to the heat capacity at constant volume. That is, $$C_p=C_v=C$$

As far as entropy is concerned, increasing the pressure on an incompressible fluid reversibly involves no heat transfer, so the entropy change is just $$\Delta S=\int{\frac{C(T')}{T'}dT'}$$Entropy of an incompressible fluid is independent of pressure.

Note that this all assumes that thermal expansion of the liquid is also negligible.

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  • $\begingroup$ Just the best possible answer as always :) $\endgroup$ – Ghartal Dec 20 '17 at 7:19
  • $\begingroup$ Sir, how do we prove that $C_v$ is independent of $v$? $\endgroup$ – Ghartal Oct 10 '18 at 17:23
  • $\begingroup$ If the fluid is incompressible, v can’t change, right? $\endgroup$ – Chet Miller Oct 10 '18 at 17:50
  • $\begingroup$ That's right sir. $\endgroup$ – Ghartal Oct 10 '18 at 18:18
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    $\begingroup$ Well, we have that $$\left(\frac{\partial U}{\partial T}\right)_v=C_v$$ and $$\left(\frac{\partial U}{\partial V}\right)_T=0$$ so $$\frac{\partial^2 U}{\partial T \partial V}=\frac{\partial^2 U}{\partial V \partial T}=\frac{\partial C_v}{\partial V}=0$$ $\endgroup$ – Chet Miller Oct 10 '18 at 19:31

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