Question

I need help with a macroscopic systems problem, concerning entropy.
This is for a quiz, so I'm hoping for only relevant information, no processes if possible, thanks!

We have two systems separated by a movable, heat-conducting partition. Thus, \(S=S(E,V)\).
In the section of my book that corresponded to this problem, it was defined just by reasoning that for a similar system without the movable partition aspect \(\dfrac{\partial S}{\partial E}\equiv\dfrac1T\).
Now it's asking for the more general case, with this movable partition.
(continued below)

Answer 1

It says that, just like \(\dfrac1T\) has a definition, \(\dfrac PT\) also has a definition.

The question asks:
(a) Why should pressure come into play, and to what might \(P/T\) be equated? (Note: Check to see whether the units make sense.)
(b) Given this relationship, show that \(dS=dQ/T\). (Remember the first law of thermodynamics.)

I know that the first law relates to
\(dE=dQ-dW\) where my professor uses \(W\) as work done by the system, apparently. (I know people have different preferences, but that's what I should use in the end.)

I think that pressure will come into play because \(dW = PdV\). Or maybe it's as simple as \(\dfrac{\partial S}{\partial E}\equiv\dfrac PT\) as entropy is proportional to pressure. I don't know!

Answer 2

In the case of an immovable partition, \(dE=dQ\) and so the \(dS=\dfrac{dQ}T\) is trivial.

Now, though, \(dE=dQ-dW\)! I know that the work done by a one system is on the other, so the change in energy of the two systems due to work is zero overall. But I don't have the relationship to see if I can use that line of thought. At least, I haven't worked it out yet.

I have to go, but I'll be back in 8-9 hours.. I have work... But I appreciate any help, thankyou!

Answer 3

thank you*

Answer 4

Well, it definitely can't be \(\dfrac{\partial S}{\partial E}\equiv\dfrac PT\) because that would imply that \(\dfrac PT\) has the same units as \(\dfrac1T\). And that's not right.

Answer 5

I'm not quite sure about the first part.
I mean, the obvious thing if it's an ideal gas in the volume, then P/T = Nk/V.

Which I guess is somewhat useful, because then P = NkT/V

You can then integrate this over a change in volume to find the work done. Temperature is not changing. That also means that the change in E is 0, since temperature is not changing.

\[\delta E=Q+W=0\]\[Q = -W=-(-PdV)=PdV\]
Now we integrate:
\[Q = \int\limits_{Vi}^{Vf}PdV=\int\limits_{Vi}^{Vf}\frac{ NkT }{ V }dV\]\[Q = NkTln(\frac{ Vf }{ Vi })\]
If you're at this point in your studies, you should already know there is an expression for the change in entropy for this process.
\[\delta S =Nkln(\frac{ Vf }{ Vi })\]
Notice, the expression for Q above is only off by a factor of T

\[Q = T \delta S\]
Hence: \[\delta S = \frac{ Q }{ T }\]

Answer 6

Thank you very much for your response!

I was considering the Ideal Gas Law also, but I'm hesitant because it wasn't explicitly mentioned when deriving \(dS=\dfrac {dQ}T\).

And thanks for the smooth derivation! Unfortunately, I don't think I've learned that \(\partial S=Nk_B\ln\left(\dfrac{V_f}{V_i}\right)\), just that \(S=k_B\ln\left(\omega\right)\) where \(\omega\) is the ensemble, I think.

Also, I think I need to account for temperature change, since there is a heat-conducting partition. I'll see if I can use the Ideal Gas Law regardless, though.

Thank you again, @Pompeii00 !

Answer 7

For a more general case than and ideal gas, I might also try to consider
\(P=\dfrac EV\)
and
\(dP= \dfrac{\partial P}{\partial E}dE+\dfrac{\partial P}{\partial V}dV\)

That has \(P(E,V)\) like \(S(E,V)\).

I'll try to remember to update this when I find or get the answer. It's due today, so I'll probably have the answer up within 2 weeks.

Answer 8

Although, \(dP\) is irrelevant, and temperature isn't explicit there.

Answer 9

So the only way to involve temperature is some assumption based on concepts, using the Ideal Gas Law, or something I haven't considered.

Answer 10

Yeah, the expression with the entropy I gave is actually derivable from the Sackur-Tetrode equation for constant temperature. I've recently studied this and had this derivation, but I wouldn't be surprised if there are many ways to define initial conditions and then derive the formula. In my notes, we specifically derived the formula you are trying to derive for a quasistatic isothermal expansion of a monatomic ideal gas. So that is why I assumed constant temperature, is because that is what the expression holds for.

Answer 11

It seems to me that the whole idea behind this question is, that when a system comes to thermodynamic equilibrium, it achieves a state of maximum entropy.

So, when two subsytems exchange energy and make their temperatures equal, what is happening is the entropy is being maximised with respect to energy exchange. In statistical physics, this is how we define temperature.

Now in the case where the subsystems can interact through a moveable conducting partition, entropy will be maximised as a function of energy and volume, leading to the additional definition of pressure (pressures equalise at equilibrium).

So you have \[\frac{ 1 }{ T }=\frac{ \partial S }{ \partial E}_{V}\] and
\[\frac{ P }{ T }=\frac{ \partial S }{ \partial V }_{E}\]

Now you can use the definition of pressure above in the first law in the form\[dE=dQ - pdV\]
to show fairly easily that dQ=TdS.

Answer 12

Okay, thank you @Pompeii00! That's pretty cool! Unfortunately, my professor wants it to be more general, so I won't be using the Ideal Gas Law.

Thank you, @ProfBrainstorm !
This is now due Wednesday, so I will work on it when I get a chance. I guessed at the \(\dfrac PT=\dfrac{\partial S}{\partial V}\) based on units, and that the discrepency between \(\dfrac1TdE\) and \(\dfrac PT\) is the volume in \(P=\dfrac EV\).

And I didn't mention it before, but I think it's implied to be at equilibrium, so \(\Delta S=0\). I'm still going to go by your methodology, @ProfBrainstorm . But I'm also going to see if I can do what my professor suggested I start with, which is to use \(S=S(E,V)\) to get
\(dS=\dfrac{\partial S}{\partial E}dE+\dfrac{\partial S}{\partial V}dV\)
where
\(dE=dQ-dW=dQ-PdV\)

But right now I have to study for another class! I'll post what I do when I get a chance!

Answer 13

Yes, exactly, dS = 0 is the condition for a maximum of S, with respect to the chosen variables, E and V. The system adjusts its internal degrees of freedom to maximise the entropy, this is pretty much the definition of equilibrium, at least in statistical physics.

Answer 14

Okay, thank you!

Answer 15

Okay! I have my solution, and I think it's correct! I'll try to remember to post if I find out that I'm wrong!

So, for this question (still learning the fundamentals),
a)
The units of \(\dfrac PT\) are equivalent to that of \(\dfrac E{VT}\) considering \(P=\dfrac EV\). Thus, we see quickly that the units of \(\dfrac PT=\dfrac E{VT}\) are \(\rm\dfrac J{m^3\dot\ K}\). This differs from the units of entropy only by the \(\dfrac1{\rm m^3}\), suggesting volume divides the entropy to equal \(\dfrac PT\).

A reasonable assumption would be that \(\dfrac PT=\dfrac{\partial S}{\partial V}\). Note that @ProfBrainstorm stated earlier that \(\dfrac{ P }{ T }=\dfrac{ \partial S }{ \partial V }_{E}\). This is more accurate as the subscript to the fraction, "\(E\)", indicates that \(E\) was considered constant for that partial derivative.

I will now use that notation.

b)
We now begin to examine \(S=S(E,V)\).
This implies that
\(dS=\left(\dfrac{\partial S}{\partial E}\right)_VdE+\left(\dfrac{\partial S}{\partial V}\right)_EdV\)

Through the definition in the book, where \(\dfrac{\partial S}{\partial E}\equiv\dfrac1T\), and through our new assumption that \(\left(\dfrac{\partial S}{\partial V}\right)_E\equiv\dfrac PT\), we can substitute usefully for the incremental change in entropy. Now,
\[dS=\dfrac1TdE+\dfrac PTdV\]

The \(PdV\) looks familiar, as it is equivalent to \(dW\) in the first law of thermodynamics. Substituting,
\[dS=\dfrac{dE}T+\dfrac {dW}T=\dfrac1T\left(dE+dW\right)\]

Seeing \(dE\) and \(dW\) together is also familiar from the first law.\[dE=dQ-dW\qquad\Rightarrow\qquad dE+dW=dQ\]

Substituting, \[dS=\dfrac1TdQ\]

\(\large\frak{Q.E.D.}\)

Answer 16

Nice work

Answer 17

Thank you, and thank you both for your help! :)