# Some basic thermodynamics

1. Oct 5, 2015

### sa1988

I've just transferred to a new university where they did certain aspects of thermodynamics in their first year, which is a problem because I didn't do any in my first year at the university I was previously at. I did some bits in high school but not to a very high level, so I understand the overall concept of what's going on, but am a little unsure/rusty on some things.

1. The problem statement, all variables and given/known data

A) Why is the ideal gas equation a better approximation for an argon gas than for a xenon gas at the same temperature and density? [2 marks]

B) i) A sample of argon expands reversibly and adiabatically to twice its initial volume. Calculate the final pressure of the gas. [3 marks]
B) ii) Calculate the final temperature of the gas if the initial temperature is 25°C. [3 marks]

2. Relevant equations

PV=nRT

3. The attempt at a solution

A)

I wrote that argon is better suited to the ideal gas equation because it as an overall smaller molecule, hence the ideal gas law holds for longer before issues such as van der Waals forces come into play, and there is less problem regarding the actual space the atoms take up, since the ideal gas law assumes the atoms take up no space at all. Hence the smaller argon atoms are better suited.

B) i)

PV=nRT

P = nRT/V

So when volume doubles to 2V, we have:

P = nRT/2V

I feel there's more to this, but I don't know what...

B) ii)

This is the bit I'm a little confused on. It's a quasistatic, adiabatic expansion, meaning no heat enters the system. All the energy goes as work, dQ = dW

So is there a change in temperature?

Part of me wants to say it stays the same, but another part feels like I should be doing more manipulation of PV=nRT

Thanks!

2. Oct 5, 2015

### PietKuip

A. The volume would probably only give you one of the two marks.

B. Why do you assume in (i) that temperature is constant?

3. Oct 5, 2015

### sa1988

In which case, I really am very unsure about where to go with the entire question...

4. Oct 5, 2015

### Staff: Mentor

How much thermodynamics did you miss? Did you learn the first law of thermodynamics? If so, please state it. The 2nd law? Do you know what the word adiabatic means? It's hard to help you and give you advice if we don't know where you're at. Are you taking thermodynamics now? If so, what is the most recent thing they covered?

5. Oct 6, 2015

### sa1988

First Law states dU = dQ + dW, that is the overall change in internal energy is equal to the change in heat plus the change in work.

Haven't covered the second law so far.

Adiabatic means no heat enters a system but work is done, that is dQ = 0 , thus dU = dW

I'm taking thermodynamics now and the most recent thing we covered is pretty much exactly what's in the questions I've given above.

We've also done a lot of calculus regarding the first law, and using it to show CP - CV = nR

It's also given that CP / CV = γ , which is one of the things I apparently missed last year.

However, aside from all this, I think the homework question is primarily focussed on recapping things we 'already know' (this has been the case in all my modules so far, since it's the very start of the semester), so the given problems should be based on more basic principles than any of the current course content.

Regardless of what I do or don't know, I'd still be keen to know the most thorough way to answer the question, from people who know exactly how to answer it. Thanks :)

6. Oct 6, 2015

### Staff: Mentor

OK. So, for part 2, dQ = 0. So, dU = nCvdT=-PdV. Does this equation make sense to you?

7. Oct 7, 2015

### sa1988

Yes, although it seems you're referring to the molar heat capacity, which is a form I've never really dealt with before.

I presume this can also be written as dU = mCvdT = -PdV ?

Still, from here I'm a little unsure of how it relates to finding the final answer of the question regarding temperature change, since the values of m and Cv are not given.

OR....

Is the answer just an algebraic rearrangement...?

mCv dT = -PdV

So... ΔT = -PΔV / mCv

And we know ΔV is 2V

So, finally, from a start temperature of 25 °C, we can say

Final temperature = 25+ΔT = (25 - 2PV / mCv)°C ....?

Last edited: Oct 7, 2015
8. Oct 7, 2015

### Staff: Mentor

Unfortunately, you have some catching up to do. I suggest you start vigorously going through your textbook and that you contact your professor so that he can help you catch up. At Physics Forums, we can't provide a complete tutorial on Thermodynamics to the extent that you currently need. It would be beyond the scope of what we do.

Chet

9. Oct 7, 2015

### sa1988

Sorry, I realised I do know what you meant with dU = nCdt = -PdV

It refers to specific heat capacity, which is something I've studied already.

I just hadn't seen it in that form before.

I've totally edited my post above which hopefully shows some progress.

10. Oct 7, 2015

### Staff: Mentor

OK. Here's a reality check:

Did you know that, for a reversible process, the first law of thermodynamics can be expressed as dU = dq - dW?
Do you know the definition of a reversible process?
Did you know that, for an ideal gas, the internal energy is a function only of temperature T?
Did you know that, for an ideal gas, PV = m RT?
Did you know that dW = PdV for a reversible process path?

See what I mean?
This is all incorrect because P is not constant. You need to substitute the pressure from the ideal gas law into the first equation.

If you look in your textbook, you will see all of this worked out in there.

Chet

11. Oct 8, 2015

### sa1988

γ
Yes
Yes. An idealised, 'quasistatic' process which take places infinitely slowly such that the system is at equilibrium at every stage.
No.
Yes
Yes

Thank you.

Ok so I spent several hours last night, and today, looking at related content in my book and have gone thought quite a lot of it and managed to work through and establish quite a few key points.

I see that the equation you gave me, dU = nCvdT=-PdV, is used in establishing the relations:
P1V1γ = P2V2γ
and
T1V1γ-1 = T2V2γ-1

I manually went through these points step by step and wrote out each stage to ensure I understand how they're derived. (I'm saying this because I worry that you think I'm just regurgitating random formulae in the book)

So, to answer the first question I can say that when the volume doubles, we have:

P2 = P1(V1/2V2)γ
P2 = P1(1/2)γ

In the first part of this homework exercise I was asked to establish the value γ for a mono-atomic particle. This gave me γ = 5/3

So in conclusion, P2 = P1(1/2)5/2

With the given information, I don't see how I can get any more accurate than that?

For the next part, we're told the starting temp T = 25°C = 298.15K

So, using T1V1γ-1 = T2V2γ-1, we say

T2 = T1(V1/2V2)γ-1
T2 = 298.15*(1/2)(5/2)-1
T2 = 298.15*(1/2)(3/2)
T2 = 105.4K
T2 = -167.7°C

The temperature reduced because internal energy was lost as work in increasing the volume.

The process in getting these answers reflects an almost exactly similar example based on adiabatic compression in my textbook. If I'm still far off then I do believe it's time for me to start worrying..!

Last edited: Oct 8, 2015
12. Oct 8, 2015

### sa1988

Sorry, I made a calculation error above.

I accidentally went from γ=5/3 to γ=5/2
It should be γ=5/3

P2 = P1(2)(-5/3) = 0.31P1
T2
= 298.15*(2)-2/3 = 187.82K = -85.33°C