Isothermal Compressibility always positive proof

I have a question on the quantity -(dV/dP)T,N where V = volume, P = pressure, T = temp, N = number of moles and T, N are held constant. I see in text books that this quantity is always positive at equilibrium. It makes intuitive sense, as if it were negative, it would be unphysical. I've been playing around with different maxwell relations, partial derivate relations, but can't seem to get to it. I know that Cv/T = (dS/dT)V,N where S = entropy is a positive quantity from the 2nd law. I should be able to relate this to my problem above. I have similar issues with -(dV/dP)S,N, (dN/du)S,N where u = chemical potential, (dN/du)T,V, which are all supposed to be positive. Thanks.
 
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Hello and welcome skateboarding.

Try this elementary proof for a perfect gas, I have dropped the N as unneccessary.

Perfect Gas Law

[tex]PV = RT[/tex]

Rearrange

[tex]P = RT{V^{ - 1}}[/tex]

Partial Differentiate with respect to volume

[tex]{\left( {\frac{{\partial P}}{{\partial V}}} \right)_T} = - RT{V^{ - 2}} = \frac{{ - PV}}{{{V^2}}} = - \frac{P}{V}[/tex]

Substitute into definition of isothermal elasticity (= 1/compressibility)

[tex]{K_T} = - V{\left( {\frac{{\partial P}}{{\partial V}}} \right)_T} = - V\left( { - \frac{P}{V}} \right) = P[/tex]

So unless you can have negative pressure the isothermal elasticity, which is equal to the pressure, is always positive.

go well
 
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Yes, this is true for ideal gases, and probably in general that the negative of the compressibility is positive. But this is based on empirical evidence of positive pressure. Of course there are meta stable systems with negative pressure, but not for true equilibrium. This seems good enough justification, but somewhat unsatisfying not being able to derive the results from first principles(laws of thermo). I guess what I'm really asking is why or how the derivate of a thermodynamic variable with respect to it's conjugate has a definite sign. T being the conjugate of S, V for P, N for u. If there is no other answer than it being so from empirical evidence, then I can deal with that, much in the same way that I see the laws of thermodynamics to be true.
 
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Perhaps you might explain further?

In particular lay out where you want to start from and your objective, even if you can't fill in the intermediate steps.
 
I think I figured out the problem. Hopefully I didn't make any huge mistakes.

Consider an isolated volume partitioned by a moveable impermeable membrane.

[tex] \LARGE V_{total} = V_A + V_B[/tex]

since total volume is constant

[tex]\LARGE (1) \hspace*{5mm} dV_A = -dV_B [/tex]

We can write the total entropy of the system in terms of volume and energy, ignoring material exchange.

From the fundamental equation for entropy

[tex] \LARGE dS = \left(\frac{\partial S}{\partial U}\right)_VdU + \left(\frac{\partial S}{\partial V}\right)_UdV [/tex]

And condition of equilibrium [tex] \LARGE dS = 0 [/tex]

[tex] \LARGE (2) \hspace{5mm} dS = \left(\frac{\partial S}{\partial V_A}\right)_UdV_A + \left(\frac{\partial S}{\partial V_B}\right)_UdV_B + \left(\frac{\partial S}{\partial U_A}\right)_VdU_A +\left(\frac{\partial S}{\partial U_B}\right)_VdU_B = 0[/tex]

Since the system is isolated,

[tex] \LARGE dU_A = -dU_B \hspace*{10mm}[/tex] and also, [tex]{\hspace*{10mm} \LARGE \frac{1}{T} = \left(\frac{\partial S}{\partial U}\right)_V [/tex]

Plugging this into [tex] \LARGE (2) [/tex] we get

[tex] \LARGE (3) \hspace{5mm}dS = \left(\frac{\partial S}{\partial V_A}\right)_UdV_A + \left(\frac{\partial S}{\partial V_B}\right)_UdV_B + \left(\frac{1}{T_A} - \frac{1}{T_B}\right)dU_A = 0[/tex]

[tex] \LARGE T_A = T_B [/tex] since we are only considering mechanical equilibrium

[tex] \LARGE (3) [/tex] reduces to

[tex] \LARGE (4) \hspace{5mm}dS = \left(\frac{\partial S}{\partial V_A}\right)_UdV_A + \left(\frac{\partial S}{\partial V_B}\right)_UdV_B = 0[/tex]

Now using the identity [tex]\hspace*{10mm} \LARGE \frac{p}{T} = \left(\frac{\partial S}{\partial V}\right)_U \hspace*{10mm}[/tex] and [tex]\hspace*{5mm}(1) [/tex]

[tex] \LARGE (5) \hspace{5mm} dS = \left(\frac{P_A - P_B}{T}\right)dV_A = 0[/tex]

At equilibrium [tex] \LARGE P_A = P_B [/tex]

We can write [tex] \LARGE (5) [/tex] as

[tex] \LARGE (6) \hspace{5mm} \frac{dS}{dV_A} = \left(\frac{P_A - P_B}{T}\right)[/tex]

Now for the tricky part

Suppose the system deviates from equilibrium slightly, and [tex] \LARGE P_A > P_B [/tex]

From [tex] \LARGE (6) [/tex] we see that the derivative is positive. Since always, [tex] \LARGE dS > 0 \hspace*{10mm} A[/tex] must have a positive displacement [tex] \LARGE \Delta V_A > 0 [/tex]

At the same time, the condition for equilibrium dictates that [tex] \LARGE P_A [/tex] will decrease on it's approach to equilibrium. [tex] \LARGE \Delta P_A < 0 [/tex]

Taking this all together,

[tex] \LARGE \frac{\Delta V}{\Delta P} < 0 \hspace*{10mm}[/tex] and for an infitesimal change [tex] \hspace*{10mm} \LARGE \frac{dV}{dP} < 0 [/tex]

From this we see that

[tex] \LARGE -\left(\frac{\partial V}{\partial P}\right)_{T,N} > 0 [/tex]

Where the term N is the number of molecules, since we were looking at only mechanical equilibrium

There's probably a more elegant solution, but I think this works. This only assumes the first and 2nd laws of Thermodynamics. A similar argument for the conjugates S, T and N, u should get a similar answer.

Once the result is obtained for [tex] \LARGE -\left(\frac{\partial V}{\partial P}\right)_{T,N} [/tex] a series of maxwell relations and Cp will lead you to the sign of[tex]\LARGE-\left(\frac{\partial V}{\partial P}\right)_{S,N} [/tex]

I will post the derivation if anyone is interested.
 

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