# Homework Help: Thermal Physics PV problem

1. Jan 22, 2012

### Squires

Okay, so firstly sorry if this is a poor post/wrong topic, I'm kind of new here, and it's been a while!

I'm given an example of a quasistatic process, in a frictionless piston. The piston compresses an ideal gas from Vi to Vf, and pressure increases from Pi to Pf, all at a constant temperature.

I've calculated the final pressure, work done etc, but I'm now asked to calculate the internal energy of the gas(ideal), and then the heat flow into the system, and I'm getting confused.

Attempted solutions:

Obviously if the temperature is constant, then heat transferred(Q) to the system is 0, and so heat flow in must equal heat flow out surely?

so from the first law of thermodynamics, the change in U(int energy)=Q+W(workdone),

so the change in internal energy is equal to work done? But then how do I calculate the heat flow into the system?? I dont think its 0, because I require the heat flow in the next question, which is to calculate the entropy change, which is just deltaQ/Temperature if I'm correct?

Thanks for reading, sorry if it came across hard to understand what I'm trying to say, explaining science is definitely not my forté :S

Last edited: Jan 22, 2012
2. Jan 22, 2012

### Squires

Or do I use the enthalpy equation dH=dQ + VdP, calculate the dH by calculating H for each ViPi and VfPf, and then dQ is the heat flow?

3. Jan 22, 2012

### rude man

4. Jan 22, 2012

### Squires

Okay, so that means that there is no change in internal energy? As, theoretically all the energy that would have been gained is lost due to the free flowing heat leaving the piston?

giving dQ=dW,

but I thought the 1st law was expressed as dU=dQ+dW, not minus?

so dQ=-dW? which makes more sense from the reasoning of dU being 0?

5. Jan 22, 2012

### rude man

Yes, the heat gained by pushing the piston from Vi to Vf is transferred to the temperature reservoir T. Note that there is no Q flowing into the gas, just out of it. W changes to Q within the gas. All of it.

Right.

Conventionally, it's dU = dQ - dW. Historically, the first app for this equ'n was steam engines, where W was considered positive. In your case dW is negative since work is being done ON the system rather than BY it. W is negative and Q is also negative (heat LEAVES the system & flows into the reservoir).
No it doesn't, not if W is conventionally defined.

You have to keep in mind how the signs of Q and W are defined w/r/t your system (gas).

There are apparently a few oddballs who do consider W positive if work is done ON the system. Just forget that, you've got enough on your plate as it is. If your book says dU = dQ + dW, get another book! If your instructor says dU = dQ + dW, get another instructor! And so on.

Ref: E. Fermi, Thermodynamics, Dover 1936

6. Jan 22, 2012

### Squires

Yesss this makes so much sense now, I had this theory earlier, but had convinced myself that the internal energy would have to change, and because of the constant temperature there was equal heat flow, but this makes so much more sense, thankyou so much!

The question is actually for for an exam I'm taking on tuesday, we get a copy of the exam without the values inserted etc, so thanks so much man, means alot!

As for the -dW, it definitely makes sense that you are doing work onto the system, forcing compression, and not the system doing work. In his lecture notes it's a +, and he wrote the exam, with the question worded 'calculate the heat flow into the system', but I'm going to go with -, explain myself out, his a sound guy and I doubt he'll dock me marks if I explain myself thoroughly.

Thanks again man, have a chilled sunday

7. Jan 22, 2012

### rude man

Oh yeah, I know the instructor is king. I have no problem with him saying that positive work is done on the system, which of course it is, then calling that W. It's just that when you deal with more general situations you want to keep a consistent sign convention, and the way that's done almost by everyone is to say W is positive when done by a steam engine's system, which everyone understands. It's just to avoid confusion when corresponding with others. And I get a warm & fuzzy feeling being on the same team with Enrico Fermi, Mark Zemansky and the other big boys ...

Cheers!

8. Jan 22, 2012

### Redbelly98

Staff Emeritus
About the ΔU = Q ± W thing...

Both are used. What I have read about it is: engineers are mostly interested in the work done by the system, eg. in a heat engine, so they define W to be positive when work is done by the system on the surroundings. On the other hand, chemists are mostly interested in the system itself, so they define W to be positive when work is done on the system.

There's not much point in arguing over which definition is correct or better than the other.

9. Jan 22, 2012

### rude man

Didn't know that. Well OK, will have to keep it in mind for future reference.
No argument desired.
I just assume my targets are physics and engineering students rather than chemists, per the two forums I goof around in. I never enter the "Other Sciences" forum where the chemists reputedly reside.

10. Jan 23, 2012

### Squires

Yeah I went back and looked over the work thing aswell, just to be sure. It appears that work done is negative when the system does work on teh surroundings, and the dU=dQ+dW is still correct, it's just when we calculate our work, it's calculated as negative, so dU=dQ+-dW -> dU=dQ-dW for my equation, which, as a physicist, I am more than happy to accept logically.

I don't know if you guys are us or uk, but here in uk I notice alot of differences in thermal physics to us. The most annoying has to be the carnot engine, Qc=heat removed from cold resovoir, and Qh=heat supplied to hot resovoir, whyyyy why why why why, seriously, just keep them both to supplied or both taken!

11. Jan 23, 2012

### rude man

I say, rum luck! Seriously, I prefer T1 and T2 where T1 is always > T2. But I see nothing egregious about using Tc and Th either.

I'm not sure I'd prefer "supplied" and " taken". Why would that be better?

Squires, as a Brit, are you upset that the SI unit of energy is named after a Frenchman and not the BTU?

Hey, no offence meant, I have great admiration for the UK and in fact, could I afford it, I'd probably be living in London right now, and not in Phoenix, AZ USA! Pardon my ramblings.

Cheers!

12. Jan 23, 2012

### Squires

Ahh yeah totally, T1 and T2 any day, but I had a question to derive it from Q1 and Q2, and made a schoolboy error of using wiki, to only find out Q1 and Q2 had opposite direction, physicsrage!

Haaaahaha, I'm not loosing sleep over it just yet, don't worry ;)

And as for London, believe me it doesn't matchup to the postcards buddy!

Peace!

13. Jan 24, 2012

### Telemachus

As callen defines it dU=dQ+dW, the work is positive when it's done over the system, so the system gains internal energy, so then he defines dW=-PdV, and thats the positive work done on the system. If we think of a piston sliding inside a cylinder, as when we compress it, and work is done on the system, we have a negative dV, so the minus signs generates positive work.

14. Jan 24, 2012

### rude man

And who be callen?

15. Jan 24, 2012

### vela

Staff Emeritus
The guy who wrote this book. The first few chapters contain an amazingly clear exposition of basic thermodynamics.