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1st Law Of Thermodynamics

by eliassiguenza
Tags: thermodynamics
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eliassiguenza
#1
Apr1-11, 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=Q-W (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!
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Doc Al
#2
Apr1-11, 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.
Penn.6-5000
#3
Apr3-11, 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 such-and-such 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 issue---conservation of energy---you're generally better off making it up as you go along.

netheril96
#4
Apr4-11, 08:33 AM
P: 193
1st Law Of Thermodynamics

Quote Quote by Doc Al View Post
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.
My textbook says U=Q+W is the engineering convention, so we physicists adopt it too.
Naty1
#5
Apr4-11, 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.
netheril96
#6
Apr4-11, 09:17 AM
P: 193
Quote Quote by Naty1 View Post
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.
Gravitational potential energy being negative is totally different with this. You may change the definition of W so its sign reverses, but you can't change the definition of potential energy, or total energy would no longer be conserved.
ashishsinghal
#7
Apr14-11, 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.
Andrew Mason
#8
Apr14-11, 06:32 PM
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Quote Quote by eliassiguenza View Post
Physicist Say:

U=Q+W (When a system is expanded the gas does work so it looses Q therefore is negative)

Engineers Say:

U=Q-W (When a system is expanded the gas does work on the outside so therefore is positive)
The convention for W has changed over the last several decades.

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
eliassiguenza
#9
Apr14-11, 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... !
Doc Al
#10
Apr15-11, 06:32 AM
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Quote Quote by Andrew Mason View Post
The convention for W has changed over the last several decades.

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]
Interesting. I would have said just the opposite. My 1966 Halliday and Resnick used [itex]\Delta U = \Delta Q - \Delta W[/itex], but a more modern text such as Randy Knight's 2004 book (picked at random) uses [itex]\Delta U = \Delta Q + \Delta W[/itex].

(Not that it matters much. Or that I've done a systematic survey.)
Andrew Mason
#11
Apr15-11, 12:46 PM
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Quote Quote by Doc Al View Post
Interesting. I would have said just the opposite. My 1966 Halliday and Resnick used [itex]\Delta U = \Delta Q - \Delta W[/itex], but a more modern text such as Randy Knight's 2004 book (picked at random) uses [itex]\Delta U = \Delta Q + \Delta W[/itex].

(Not that it matters much. Or that I've done a systematic survey.)
I may have overstated that bit. I should have said that I had the impression that the conventional statement was dU = dQ + dW (or dQ = dU - dW). I am just going by my Zemansky text from the early 70s.

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 old-school 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
eliassiguenza
#12
Apr16-11, 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!?
Doc Al
#13
Apr16-11, 05:30 AM
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Quote Quote by eliassiguenza View Post
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!?
Right.
Pythagorean
#14
Apr16-11, 05:36 AM
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Quote Quote by eliassiguenza View Post
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!?
only if you're talking about work done by the system
Doc Al
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Apr16-11, 06:24 AM
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Quote Quote by Pythagorean View Post
only if you're talking about work done by the system
Nope. W only stands for the work done by the system in the second version (the one with the minus sign).
Pythagorean
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Apr16-11, 12:30 PM
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Wait, so this statement would be wrong?:

If U = Q + W, Work done on the system is positive

If U = Q - W, Work done on the system is negative
but wouldn't this be equivalent to:

If U = Q + W, Work done by the system is negative

If U = Q - W, Work done by the system is positive
(I'm assuming W is always a positive value in the above and the sign tells the direction the energy is "going", is that my mistake?)
Doc Al
#17
Apr16-11, 02:20 PM
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Quote Quote by Pythagorean View Post
(I'm assuming W is always a positive value in the above and the sign tells the direction the energy is "going", is that my mistake?)
Yes, that's your mistake. In one version, W stands for the work done on the system; in the other, it stands for the work done by the system. In one case W is negative; in the other, positive.
Pythagorean
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Apr16-11, 04:13 PM
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Thanks, Doc.


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