Computing entropy change of ideal gas during heating

[jasonRF]

Homework Statement

Compute the entropy change of n moles of an ideal gas that is heated from $T_i$ to $T_f$ at constant volume. You know the heat capacity of this gas is $C_p(t)$.

Relevant Equations

For an ideal gas $$C_p - C_v = nR$$.

This is my first post of a homework problem, and I am just trying to make sure I am not missing anything as I help a child who is taking a college class on this stuff this summer. And to be clear, just trying to help her understand stuff, not help her with the homework problem. So I won't go through all the steps or give the explicit equation for $C_p(t)$ or anything.

To me this seems like it should be very straightforward. Since this is a constant volume process, $\delta q = C_v(T) dT = (C_p(T)-nR) dT$, from which we can get dS and integrate. Am I missing something? I am mainly asking because the TA indicated to my child that they should just use $\delta q = C_p(T) dT$ since entropy is a state function, which makes no sense to me nd was of course very confusing to my kid (who also may have misunderstood the TA - I of course wasn't there).

But I am not confident about classical thermo since I don't know much am mostly self-taught so I don't want to "correct" what my kid heard from the TA without checking with folks here who know this stuff better than I do. Any enlightenment would be much appreciated.

Thanks,

Jason

 

[kuruman]

I would use the first law in the form $dU=T~dS-p~dV$. What happens to the terms in it when the volume is kept constant?

 

[vela]

Homework Statement: Compute the entropy change of n moles of an ideal gas that is heated from $T_i$ to $T_f$ at constant volume. You know the heat capacity of this gas is $C_p(t)$.
Relevant Equations: For an ideal gas $$C_p - C_v = nR$$.

Typically, $C_p$ and $C_V$ are molar heat capacities, so $C_p - C_V = R$. You may want to check what convention in being used.

To me this seems like it should be very straightforward. Since this is a constant volume process, $\delta q = C_v(T) dT = (C_p(T)-nR) dT$, from which we can get dS and integrate. Am I missing something? I am mainly asking because the TA indicated to my child that they should just use $\delta q = C_p(T) dT$ since entropy is a state function, which makes no sense to me nd was of course very confusing to my kid (who also may have misunderstood the TA - I of course wasn't there).

Your approach sounds fine.

Is the heating at constant volume the entire problem or is it one leg of a two-step process? If it's the former, then what the TA said doesn't make sense. But if the problem involves allowing the gas to expand after it's heated until it reaches its initial pressure, then you could calculate the entropy change for the entire process using an isobaric expansion since entropy is a state function. I give a problem like this to students to do in class when learning how to calculate entropy changes.

 

[jasonRF]

Typically, $C_p$ and $C_V$ are molar heat capacities, so $C_p - C_V = R$. You may want to check what convention in being used.Your approach sounds fine.

Is the heating at constant volume the entire problem or is it one leg of a two-step process? If it's the former, then what the TA said doesn't make sense. But if the problem involves allowing the gas to expand after it's heated until it reaches its initial pressure, then you could calculate the entropy change for the entire process using an isobaric expansion since entropy is a state function. I give a problem like this to students to do in class when learning how to calculate entropy changes.

Thanks for the reply. This question is just the one-step process, which was why my daughter became so confused - to her credit she decided to leave her solution anyway (before asking me about it) instead of changing it to agree with what she understood from the TA because it made more sense to her. But it did put a lot of doubt in her mind. I just spoke with her just to assure her that she isn't crazy; she thinks it was most likely a miscommunication / misunderstanding because the TAs are very good.

I appreciate you mentioning that there might be issues with what I wrote in terms of units and whether the heat capacities were molar. In this particular case, though, I don't need or want to be too detailed so that I stay clear of any academic integrity issues. I haven't even seen my daughters' solution, since the goal on my end was just to help resolve her doubts about the concepts. I don't much care if she gets the final correct answer.

Cheers!

Jason

 

[kuruman]

I don't much care if she gets the final correct answer.

Surely you care that she gains some physical understanding from this experience. The straightforward path relies on the first law of thermodynamics and definitions. It is the suggested approach by the TA who, apparently, provided just a formula without explaining its provenance. Here is the background.
From the first law of thermodynamics
$dU=dQ-p~dV.$
At constant volume, the case here,
$dU=C_v~dT~$ and $p~dV = 0.~$ Furthermore, $~dS=\dfrac{dQ}{T}\implies dQ=~T~dS.$ Putting all this in the first law equation :
$C_v~dT=T~dS-0\implies dS=\dfrac{C_v~dT}{T}.$

 

[Chestermiller]

Tell the TA to go out and learn thermo before he teaches your daughter.

 

[jasonRF]

Surely you care that she gains some physical understanding from this experience. The straightforward path relies on the first law of thermodynamics and definitions. It is the suggested approach by the TA who, apparently, provided just a formula without explaining its provenance. Here is the background.
From the first law of thermodynamics
$dU=dQ-p~dV.$
At constant volume, the case here,
$dU=C_v~dT~$ and $p~dV = 0.~$ Furthermore, $~dS=\dfrac{dQ}{T}\implies dQ=~T~dS.$ Putting all this in the first law equation :
$C_v~dT=T~dS-0\implies dS=\dfrac{C_v~dT}{T}.$

Yes, I care that she learns the material, and I believe she basically did what you are describing. She usually starts from first principles for every problem whenever possible since it helps her understand what is going on, and she has a hard time remembering the conditions under which other derived formulas apply anyway. When the professor posted the equation sheet they are allowed to use on exams she almost laughed, since she re-derives a fair number of the given equations over and over as needed in the course of solving problems.

The issue was that she thought (and she may have misunderstood) that the TA said to essentially use that formula but with the given equation for $C_p(T)$ instead of $C_v(T)$. I was almost sure that was wrong, but she was less sure. And this was supposed to be an easy problem on this assignment; the others involved free energies, mixtures, phase diagrams, Maxwell relations, fugacity, etc., so having doubts about the easiest problem was making her doubt all the other material as well.

This is the second semester of a physical chemistry sequence; the first semester was quantum mechanics and it was pretty tough for her so she has less confidence in herself than usual. Fortunately, thermodynamics seems to be a little easier for her because she can have some physical understanding before jumping into the math. Sprinting through two semesters of material in 8 weeks (with 12 lecture hours a week) over the summer doesn't seem to be a great recipe for learning. At least the labs are in-person only during the regular Fall and Spring semesters.

Thanks!

jason

 

[kuruman]

$\dots~$since she re-derives a fair number of the given equations over and over as needed in the course of solving problems.

That's the proper way to do it. Fewer things to memorize and one gets a coherent picture in one's hed about how physical entities are interconnected.

$\dots~$so having doubts about the easiest problem was making her doubt all the other material as well.

And here, IMO, is the lesson to be learned that your daughter can use as a confidence-builder. In physics it's not a matter of opinion; when you're right you're right and when you're wrong you're wrong. The derivation of the answer in post #5 shows that the TA's suggestion was wrong and @Chestermiller's suggestion in post #6 shows his confidence in formulating it and being right. Your daughter seems to be on the path to be confident when she's right. She just needs to encounter a few more situations like this where others are demonstrably wrong while she knows she's right.

 

[Chestermiller]

To determine the entropy change of a system between an initial thermodynamic equilibrium state and a final thermodynamic equilibrium state using fundamentals, one must first devise a reversible process between these same two end states. Then one must determine the integral of dQ/T for that reversible process path. Since your daughter likes to use fundamentals rather than memorized formulas, what is the reversible process she would use to bring about the change of one mole of ideal gas from $T_i$ to $T_f$ at constant volume?