Thermodynamics question from Callen

[Ionophore]

Hi all,

I'm working through Callen to try and teach myself some thermodynamics before the semester starts and I have to take pchem. I'm pretty stumped on problem 4.2-4.

Homework Statement

A system obeys the eqns:
T=A(v^2)/s
P=-2Avln(s)
(A is some positive constant)

The system undergoes free expansion from v_0 to v_f. Find the final temperature in terms of t_initial, v_0 and v_f. Find also the increase in molar entropy.

The Attempt at a Solution

I've fumbled around a bit. I found the "fundamental relation" by integrating (by inspection) du=Tds - Pdv, it's:
U=A(v^2)ln(s)
But this leads me nowhere.

The only other thing I can see to do is say that in a "free expansion," du=0, and therefore Tds = Pdv. I can substitute into this the equations for T and P given in the problem and integrate. I end up with something ugly and it's not clear to me where to go from there. Integration gives:

(1/2)ln(s)^2 = 2ln(v)

Any help appreciated,

-Ben

 

[olgranpappy]

Hi all,

I'm working through Callen to try and teach myself some thermodynamics before the semester starts and I have to take pchem. I'm pretty stumped on problem 4.2-4.

Homework Statement

A system obeys the eqns:
T=A(v^2)/s
P=-2Avln(s)
(A is some positive constant)

The system undergoes free expansion from v_0 to v_f. Find the final temperature in terms of t_initial, v_0 and v_f. Find also the increase in molar entropy.

The Attempt at a Solution

I've fumbled around a bit. I found the "fundamental relation" by integrating (by inspection) du=Tds - Pdv, it's:
U=A(v^2)ln(s)
But this leads me nowhere.

Actually, you found, by inspection
$$du = d(Av^2\log(s))$$
which means that
$$u=Av^2\log(s) + C$$
where C is a constant.

The only other thing I can see to do is say that in a "free expansion," du=0, and therefore Tds = Pdv.

Actually, I believe that you are right about du=0 but free expansion is *not* a reversable process so you can't use dQ=TdS. But regardless the final macroscopic state does have a well-defined energy given by the expression you found because it could be reached by a reversable quasi-static series of steps...

So, solve for C in terms of u_0 v_0 and s_0 then replace s with s as a function of v and T via the equation for T. then you will have an equation for u_f in terms of u_0 v_0 t_0 and v_f. then set u_f = u_0 to get an equation for T_f in terms of T_0 and v_0 and v_f...

I believe that will work.

 

[Ionophore]

So as I understand it the strategy is to simply write an equation for the initial energy using the equation for U that I derived and v_i, T_i, and an equation for the final energy using v_f, T_f. Then, set these equal because in a free expansion there is no change in the internal energy.

I'm not sure what you mean when you say to solve for C. Isn't it just some standard state energy that will disappear when I set the initial and final energies equal?

u_i = A(v_i)^2log[A(v_i)^2/T_i] +C
u_f = A(v_f)^2log[A(v_f)^2/T_f] +C

I can now set u_i = u_f and solve for T_f, but I get something truly horrendous that can't possibly be right (Well, I suppose it could be right but I've never seen an equation like it).

 

[olgranpappy]

So as I understand it the strategy is to simply write an equation for the initial energy using the equation for U that I derived and v_i, T_i, and an equation for the final energy using v_f, T_f. Then, set these equal because in a free expansion there is no change in the internal energy.

I'm not sure what you mean when you say to solve for C. Isn't it just some standard state energy that will disappear when I set the initial and final energies equal?

u_i = A(v_i)^2log[A(v_i)^2/T_i] +C
u_f = A(v_f)^2log[A(v_f)^2/T_f] +C

I can now set u_i = u_f and solve for T_f, but I get something truly horrendous that can't possibly be right (Well, I suppose it could be right but I've never seen an equation like it).

Yeah, that's right. You get
$$v_i^2\log(av_i^2/T_i)=v_f^2\log(Av_f^2/T_f)$$
or
$$T_f=Av_f^2e^{-\frac{v_i^2}{v_f^2}\log(Av_i^2/T_i)}$$

 

[Andrew Mason]

Homework Statement

A system obeys the eqns:
T=A(v^2)/s
P=-2Avln(s)
(A is some positive constant)

The system undergoes free expansion from v_0 to v_f. Find the final temperature in terms of t_initial, v_0 and v_f. Find also the increase in molar entropy.

If it is a free expansion (no work done, no heat flow), wouldn't the first law of thermodynamics require U, hence T, to be constant?

AM

 

[Ionophore]

"If it is a free expansion (no work done, no heat flow), wouldn't the first law of thermodynamics require U, hence T, to be constant? "

But most generally, must U depend on T?

 

[olgranpappy]

If it is a free expansion (no work done, no heat flow), wouldn't the first law of thermodynamics require U, hence T, to be constant?

AM

It's not an ideal gas. As you can see from the fundamental relation that was derived the energy depends on both the volume and the temperature, not just the temperature.

 

[Ionophore]

Right, in this case u depends on both v and T. But in general, must u depend on T? This is something that I have wondered for a very long time. It seems intuitively obvious that u should always increase with T, but I'm not sure I can nail down any bit of the theory thus far developed in Callen (I'm at about page 100) that says u must always be an increasing function of T...

 

[olgranpappy]

I can't remember a general proof at the moment, but I can cook up a cute example for you where, under a very specific set of circumstances, $$dU/dT<0$$. I.e., a temperature increase leads to an energy decrease.

But the example relies on the presence of an external field (which are legitimate entities in thermodynamics, but are not often considered)...

Anyways. Consider an icicle hanging from the top of a very tall building. Now let the temperature of the icicle *increase* until it is higher than 0 celsius. The icicle will then be melted and thus will have fallen all the way down (thus decreasing it's gravitational potential energy) to the ground.

It should be easy to counteract any gain in thermal energy of the water by making the height of the building (and thus the loss of gravitational potential energy) big enough so
that we have
$$\frac{dU}{dT}<0\;.$$

 

[Ionophore]

Very nice example! I like it!

And thanks for all the help everyone.

-Ben

 

[Ionophore]

...

On second thought, maybe I don't like it. As the icicle falls, its gravitational potential energy is converted to kinetic energy, so it doesn't lose anything. When it hits the ground it will either dump that kinetic energy as work (perhaps it lands on a paddle wheel) or heat. Regardless, dU is unchanged until we allow the system to surrender its kinetic energy as work or heat.

Overall then dU = C(T)dT.

 

[Andrew Mason]

I can't remember a general proof at the moment, but I can cook up a cute example for you where, under a very specific set of circumstances, $$dU/dT<0$$. I.e., a temperature increase leads to an energy decrease.

But the example relies on the presence of an external field (which are legitimate entities in thermodynamics, but are not often considered)...

Anyways. Consider an icicle hanging from the top of a very tall building. Now let the temperature of the icicle *increase* until it is higher than 0 celsius. The icicle will then be melted and thus will have fallen all the way down (thus decreasing it's gravitational potential energy) to the ground.

It should be easy to counteract any gain in thermal energy of the water by making the height of the building (and thus the loss of gravitational potential energy) big enough so
that we have
$$\frac{dU}{dT}<0\;.$$

But so long as Temperature, T, is a measure of the average kinetic energy of the molecules of the system at thermal equilibrium, and so long as Internal Energy, U, is defined as the total energy associated with the random, disordered motion of molecules at thermal equilibrium, they are necessarily proportional.

If you want to include potential energy of the molecules of the system in a field (eg. gravity) as part of the internal energy of the system, you are redefining the meaning of the term.

AM

 

[Ionophore]

It is correct to include the potential energy of the icicle as part of the system's internal energy. T being a measure of the average KE of the molecules in the system is a statistical mechanical idea that is absent from classical thermodynamics - to the best of my understanding.

 

[Andrew Mason]

It is correct to include the potential energy of the icicle as part of the system's internal energy.

Then you are redefining internal energy so that it is no longer internal. If you do that, internal energy becomes a relative term - the internal energy of the system must be given with respect to some external body. That is not the thermodynamic definition of internal energy.



AM

 

[Ionophore]

Yes, the internal energy is relative. According to Callen (page 12):

"Only differences in energy, rather than absolute values of the energy, have physical significance, either at the atomic level or in macroscopic systems. It is conventional therefore to adopt some particular state of a system as a fiducial state, the energy of which is arbitrarily taken as zero. The energy of the system in the fiducial state, is then called the thermodynamic internal energy of the system in that state and is denoted by the symbol U."

 

[Andrew Mason]

T being a measure of the average KE of the molecules in the system is a statistical mechanical idea that is absent from classical thermodynamics - to the best of my understanding.

It doesn't matter that physicists did not understand the meaning of temperature until the advent of statistical mechanics. Temperature may have been measured by the amount of expansion of some substance, such as mercury. This expansion, as it turns out, is proportional to the average kinetic energy of those mercury molecules over a narrow temperature range (ie the range of the thermometer). So one could argue that temperature has ALWAYS been a measure of the average KE of the molecules of a system, whether or not those doing the measuring understood what it really was.

The fact is, however, that in thermodynamics temperature of a system IS DEFINED as the average KE of the molecules of the system.

AM

 

[Ionophore]

"The fact is, however, that in thermodynamics temperature of a system IS DEFINED as the average KE of the molecules of the system."

I don't want to argue too much about this, although I am not sure I'm ready to accept that idea outright. One concern I have is that in isolated spin systems it is possible to achieve negative temperatures (below 0 K). I simply do not know enough stat mech to develop an argument either way.

But whether or not we really are or really aren't measuring the kinetic energy of molecules, my question is, given only the classical theory of thermodynamics (for example, as developed by Callen), must U increase with T. I've never heard anybody give a compelling reason that it must.

 

[Andrew Mason]

Yes, the internal energy is relative. According to Callen (page 12):

"Only differences in energy, rather than absolute values of the energy, have physical significance, either at the atomic level or in macroscopic systems. It is conventional therefore to adopt some particular state of a system as a fiducial state, the energy of which is arbitrarily taken as zero. The energy of the system in the fiducial state, is then called the thermodynamic internal energy of the system in that state and is denoted by the symbol U."

This is a correct statement by Callen. But it does not suggest that the internal energy of a system is some arbitrary value depending on its position relative to an external body (ie gravitational potential).

U is a measure of the internal potential and kinetic energy of the molecules. External potential energy does not enter into U. If the system changes its gravitational potential energy, then work is being done by or on the system.

U depends on the total energy acquired by the molecules of the system in moving from absolute zero to its temperature T. dU = nC_vdT is true by definition. But since heat capacities change (due largely to intermolecular forces), C_v is not necessarily constant for all T so U is not necessarily a linear function of T throughout all T. This is particularly apparent with a change of state.

My original question was that if there is a free expansion of the system, doesn't this necessarily mean that U does not change? And if U does not change, how can T change? It was a question.

AM

 

[Ionophore]

Is it valid to use $$dU=C_vdT$$ in such a problem? The volume is changing, so how can we use $$C_v$$?

But perhaps we are near the answer. For U to be independent of temperature we would require $$C_v =0$$. This would require $$T(dS/dT)_v = 0$$. So it appears that U is independent of T when T=0, but when T != 0, we would need to have $$(dS/dT)_v = 0$$. Can it be shown that this is illegal?

 

[Andrew Mason]

Is it valid to use $$dU=C_vdT$$ in such a problem? The volume is changing, so how can we use $$C_v$$?

For an ideal gas, dU = nCvdT even if volume changes. dQ/dT = dU/dT + dW/dT = nCv + PdV/dT

AM

 

[olgranpappy]

...

On second thought, maybe I don't like it. As the icicle falls, its gravitational potential energy is converted to kinetic energy, so it doesn't lose anything. When it hits the ground it will either dump that kinetic energy as work (perhaps it lands on a paddle wheel) or heat. Regardless, dU is unchanged until we allow the system to surrender its kinetic energy as work or heat.

Overall then dU = C(T)dT.

Okay, okay. the example is cute but flawed. Here is another one that I think is better because it exploits an amazing property of water that is different from almost every other system--On freezing water expands! In this certain circumstance
$$\frac{dV}{dT}<0$$
which is very different from most other substances. (again, here I am exploiting a weirdness of water at a phase transition).

So...

When the temperature of a full glass of water is lowered below zero Celcius the internal energy of the water decreases, yes. But the volume increases. So imagine that the system under consideration is not just the full glass of water, but the full glass of water plus a heavy weight placed on top of the glass.

The expansion of the water as T decreases will lift the weight, thus for a heavy enough weight the decrease in the internal energy of the ice will be counteracted by the increase in gravitational potential energy of the weight.

As usual, this example is rather convoluted and exploits the gravitational energy. Perhaps you can poke some holes in this example as well.

Cheers.

 

[Ionophore]

I think that that example is better. Another one that I can think of is suppose your system is a ball on a string hanging from a ceiling. If the ball expands as you raise the temperature its center of mass is lowered, and, with appropriate dimensions and expansion coefficients, I think you could work out a net energy decrease.

But Andrew made some good points. Why do they call it "internal" energy if it is dependant on some field generated by an external body? I still lean towards thinking that it is appropriate to consider such fields when calculating the internal energy. For example, in chapter 3, Callen discusses magnetic systems and he appears to do just this, although I have a very limited understanding of magenetism and I'm not sure I grasped the subtelties of that section.

 

[olgranpappy]

there are useful discussions about thermodynamics in the presence of E/M field in Landau and Lif****z volume 8 "electrodynamics of continuous media"

 

[olgranpappy]

P.S. I found a nice proof that $$C_v$$ should be positive.

One considers a system at constant temperature $$T_0$$ and volume $$V_0$$. Thus the Helmholtz free energy 'F' is a minimum. And so, any small change in the Helmholtz free energy is positive (of course, it is zero to first order, but the fact we are at a minimum rather than a maximum means the change is positive).

Thus with $$F=U-T_0 S$$ we have
$$\delta U - T_0\delta S > 0$$

but at constant volume
$$\delta U = \frac{\partial U}{\partial S}\delta S + \frac{1}{2}\frac{\partial^2 U}{\partial S^2}(\delta S)^2$$

in other words, we have
$$\frac{\partial^2 U}{\partial S^2}>0$$
since $$\frac{\partial U}{\partial S}=T_0$$

Then, in the inequality we can put the definition, $$\frac{\partial T}{\partial S}=\frac{T}{C_v}$$ and temperature is positive, thus
$$C_v>0$$

I.e.
$$\left(\frac{\partial U}{\partial T}\right)_V>0$$