Is Bernoulli's Theorem an Expression of Energy Density in Fluids?

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I think the complete thermodynamic equation should be $$\left( \frac{\partial U}{\partial V} \right)_T = -P + T \left( \frac{\partial P}{\partial T} \right)_V$$i.e. ##P = - \frac{\partial U}{\partial V}## is a special case, not applicable here
 
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etotheipi said:
i.e. P=−∂U∂V is a special case, not applicable here
Thanks for your teaching.
Thermodynamics relation says
[tex]p=\frac{F-G}{V}\neq \frac{E}{V}[/tex]
p is volume density of ##F-G##. I feel it too abstract.
 
anuttarasammyak said:
Summary:: Pressure and energy of of ideal gas are p=NkT/V, E=3/2 NkT. So p=2/3 E/V. Why it is not E/V because pressure is energy per volume? How do I reconcile this ?

As summarized.
It only works out that way for the special case of an ideal monoatomic gas.
 
As @Chestermiller noted, the 2/3 term comes from a specific case. Ultimately, I think the important thing here is that 2/3 carries no units. The fact that pressure has units of energy per unit volume implies that
[tex]p\propto \dfrac{E}{V},[/tex]
but does not in any way require
[tex]p = \dfrac{E}{V}.[/tex]
 
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Also, even in the ideal gas limit of low density, real gases do not have constant heat capacity down to absolute zero.
 
Thanks. I could confirm pressure is not internal energy density. Do we have a general interpretation what kind of energy density pressure has ?
 
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anuttarasammyak said:
Thanks. I could confirm pressure is not internal energy density. Do we have a general interpretation what kind of energy density pressure has ?
Who says it can be interpreted that way?
 
I find in Wiki Energy Density
"In short, pressure is a measure of the enthalpy per unit volume of a system."
I do not catch it. Is it helpful to understand pressure as kind of energy density ?

The article says as for magnetic field, pressure and energy density coincide. It appears so in Maxwell stress tensor.
 
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Putting one of those equations above into plain words (my plain words, if you want to shoot them down, go ahead); a static force may be described as having an incipient potential to do work, subject to it bearing on a linear displacement, and therefore a force per unit area for a given incipient displacement is the same as an energy for a given incipient change of volume.

(that is, the 'incipient energy change' is the force x 'incipient displacement')

[For example, a force of 100kN bearing on a m^2 area, per meter of displacement that force applies, it does work at the rate of 100kJ, i.e. the same work density as 100kJ/m^3.]

This only covers the first component of the thermodynamic equation above, however, it does not describe the nature of the work done to form the pressure. It is only a partial description covering the work the pressure can do.
 
Re: #3 and #10, can we say
[tex]p=\frac{H-E}{V}=\frac{G-F}{V}[/tex]
pressure as difference of the two free energy density?

For an example may I interpret that RHS constant of Bernouill's theorem
[tex]\frac{1}{2}\rho v^2 + \rho g h + p = const.[/tex]
has dimension of energy density but the value is enthalpy density (minus constant internal energy for incompressible fluids and plus kinetic and potential energy density ) ?
 
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