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1st Law Of Thermodynamicsby eliassiguenza
Tags: thermodynamics 
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#1
Apr111, 06:29 PM

P: 24

Hi to all! recently I been doing alot of thermodynamics in University, however I have a huge problem.
Because it is my first semester I have to take physics class for freshmen, however I also take at the same time physics for year 2 Uni students. (This is cuz I can, cuz I am already a Sound Engineer.) anywho........ does anynody knows what the hell with this? Physicist Say: U=Q+W (When a system is expanded the gas does work so it looses Q therefore is negative) Engineers Say: U=QW (When a system is expanded the gas does work on the outside so therefore is positive) The problem is what the hell! cuz i'm taking physics for freshman they mixed engineers with physics and now i'm confused cuz i bothered to learn for physics cuz i want to be a physicist, and now i'm getting problems wrong because of this.........> So i want to check am I right on the statements I made above? Thanks! 


#2
Apr111, 07:05 PM

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When using U = Q + W, W is the work done on the system.
When using U = Q  W, W is the work done by the system on the environment. Take your pick. The first version is becoming more popular and is the one I always use. 


#3
Apr311, 12:16 AM

P: 31

Try doing it without either equation.
I have a checking account; the total value of this account is called "V." There are four ways to change the amount of money in my account: I can write someone a check, someone else can write me a check, I can withdraw cash at an ATM, and I can deposit cash at an ATM. So, let "C" be the net value of the checks I write (subtracting out the checks that people write to me) and let "A" be the net value of the money I deposit into the ATM (subtracting out the money I withdraw). Then the change in the value of my account can be determined by [tex]\Delta V = A  C[/tex]. But there's an annoying problem: I have some friends who went to a different school, and they learned that "C" should be the total of the checks written to them minus the checks they wrote to other folks. As a result, they use a different formula: [tex]\Delta V = A + C[/tex]. So we tend to confuse each other when we talk about accounting because our formulas are similar but not quite identical. As a person on the outside, you're probably thinking, "That's a really lame thing to get confused about. When money enters my checking account, V should increase. When money leaves my checking account, V should decrease. It's so simple! You shouldn't even need a formula! Why are these guys confused? Maybe you guys would be less confused if you dropped the formulas and just used common sense instead." That's honestly how I feel about the first law of thermodynamics. A physical system has energy. There are four ways to change the energy of a system: something can do work on it, it can do work on something else, heat can flow into it, and heat can flow out of it. If heat flows into it or someone does work on it, its energy should increase. If heat flows out of it or it does work on something else, its energy should decrease. I'm honestly surprised that so many books give a formula relating U, Q, and W for what's really a simple accounting problem. Suppose you have a gas with an internal energy of 45 kJ and it's sitting in a cylinder with a piston. It's compressed in suchandsuch a way for which work is the integral of P dV, and when you evaluate that integral you get 13 kJ. Is its new energy 58 kJ or 32 kJ? One way to figure it out would be to see whether the formula you used was for work done by or work done on the gas and then use the appropriate formula with U. The other way to think about it is, "Someone compressed this gas, therefore its energy increased. The final energy must be 58kJ." Or another example: A system starts off with 12 kJ of energy. Someone does 5 kJ of work on the system by turning a crank; the system responds by doing 3 kJ of work on another object and also by losing some energy as heat. When everything's done, the system has 4kJ of energy. Question: How much energy was lost as heat? Answer: From the first law of thermodynamics, we have that 12 + 5  3  Q = 4, so we must have lost Q = 10 kJ of heat. You can jump to the numerical equation without thinking. Common sense gave us the answer more quickly than the U = Q + W equation would have. It's true that almost everywhere else in physics, it's best to take the equations at face value and use them consistently. But on this one issueconservation of energyyou're generally better off making it up as you go along. 


#4
Apr411, 08:33 AM

P: 193

1st Law Of Thermodynamics



#5
Apr411, 09:03 AM

P: 5,632

I like the checking account analogy.
+W and W in thermodynamics makes as much sense as conventional electrical current flowing in the direction of positive charge when we know the flow are electrons with negative charge. So all the arrows of current are "backwards". And why is gravitational potential energy NEGATIVE???? It's just another convention. 


#6
Apr411, 09:17 AM

P: 193




#7
Apr1411, 11:15 AM

P: 460

It is just the matter of importance. In one case they are interested on the system and in the other they are interested in the surrounding.



#8
Apr1411, 06:32 PM

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P: 6,684

Purists used to say: [itex]\Delta U = \Delta Q + \Delta W[/itex], the idea being that Q>0 and W>0 signified energy inputs and Q<0, W<0 were energy outputs. That part makes perfect sense. But the problem with that convention is that it means W < 0 if [itex]P\Delta V >0[/itex]. Substituting W for PdV becomes confusing. So now the mainstream convention (not everyone has changed) is to say that [itex]\Delta U = \Delta Q  \Delta W = \Delta Q  P\Delta V[/itex] or [itex]\Delta Q = \Delta U + \Delta W = \Delta U + P\Delta V[/itex] AM 


#9
Apr1411, 09:46 PM

P: 24

thanx guys, it is really hard to get a grip on this when i have to take two papers which include thermodynamics, and each of the teachers like to take on their own version of this.... it makes it confusing... !



#10
Apr1511, 06:32 AM

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(Not that it matters much. Or that I've done a systematic survey.) 


#11
Apr1511, 12:46 PM

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I recall how confusing it was and our professor saying that there were more and more texts using the opposite convention but that Zemansky was the best text in his opinion so that is what we used (which suggested to me that Zemansky was using the oldschool convention). It seems to me now that most authors use dU = dQ  dW but I could be wrong on that. It sure is less confusing though because you don't have to remind yourself to keep changing the sign of PdV all the time. AM 


#12
Apr1611, 03:24 AM

P: 24

so , let me get this straight .... in a situation where I have an ideal gas expanding:
If U = Q + W, Work is negative If U = Q  W, Work is positive right!? 


#15
Apr1611, 06:24 AM

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#16
Apr1611, 12:30 PM

PF Gold
P: 4,292

Wait, so this statement would be wrong?:



#17
Apr1611, 02:20 PM

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