Convention of Work in Thermodynamics

Lapetude
Messages
2
Reaction score
0

Homework Statement



I am wondering about the following conventions:

The work done by the gas is positive if the gas expands,
and negative if it is compressed.
Conversely, the work done on the gas by external force
(e.g., a moving piston) is positive for compression, and negative when the gas expands.

Why is $$ W_{A,B}=-\int_{A}^{B}PdV $$
Specifically the signs are driving me nuts in working out the total work done on the gas in the Carnot cycle.


Homework Equations



So if a piston is compressing the work done on the gas is by convention positive. But $$dV$$ is negative. So is the negative sign in $$ W_{A,B}=-\int_{A}^{B}PdV $$
a result of trying to make sure the work done is positive in compression?.

The Attempt at a Solution



Consider the Carnot cycle 1st stage of :
isothermal expansion from volume $$V_{1}$$ to volume $$V_{2}$$ at constant temperature $$T_{1}$$. If I use the above definition I get:
$$W=-\int_{V_{1}}^{V{2}}PdV=nRT_{1}Log\left(\frac{V_{1}}{V_{2}}\right)$$
However in this case $$dV$$ is positive so do I just drop the minus from the integral? And hence is the true answer:
$$W=\int_{V_{1}}^{V{2}}PdV=nRT_{1}Log\left(\frac{V_{2}}{V_{1}}\right)$$
which makes no sense if we define the work on a gas as it expands as negative. ie, the above Log will always be positive. If my first answer is correct this would imply that the total work on the gas in the Carnot cycle is negative, while the work done by the gas is positive. Surely this makes sense as we are converting heat into the external work? Thanks in advance.
 
Physics news on Phys.org
Sorry I just found
https://www.physicsforums.com/showthread.php?t=306228
which is what I'm after.
fysics4fun made it clear:
"mpkannan makes it clear- you must look at everything in context before deciding the signs on everything. A question like "what is the sign on work" is too vague.

Two statements of the First Law of Thermo seen in textbooks:

dU = Q + W and dU = Q - W

W in the first statement is "work on the gas"
W in the second statement is "work by the gas"

And so, for example, when a gas expands W would be negative in the first statement (work done on the gas is negative since positive work is being done on the surroundings).
BUT
when a gas expands W would be positive in the second statement (work done by the gas is positive since work is being done on the surroundings).

The fact that work is negative in one case but positive in the other (both for the same scenario) is not a contradiction, since in one case we're talking about work ON and in the other we're talking about work BY. Textbooks should present BOTH statements of this first law, CLEARLY defining that the first statement is work ON and the second is work BY, in my opinion."
 
Lapetude said:
Sorry I just found
https://www.physicsforums.com/showthread.php?t=306228
which is what I'm after.
fysics4fun made it clear:
"mpkannan makes it clear- you must look at everything in context before deciding the signs on everything. A question like "what is the sign on work" is too vague.

Two statements of the First Law of Thermo seen in textbooks:

dU = Q + W and dU = Q - W

W in the first statement is "work on the gas"
W in the second statement is "work by the gas"

And so, for example, when a gas expands W would be negative in the first statement (work done on the gas is negative since positive work is being done on the surroundings).
BUT
when a gas expands W would be positive in the second statement (work done by the gas is positive since work is being done on the surroundings).

The fact that work is negative in one case but positive in the other (both for the same scenario) is not a contradiction, since in one case we're talking about work ON and in the other we're talking about work BY. Textbooks should present BOTH statements of this first law, CLEARLY defining that the first statement is work ON and the second is work BY, in my opinion."
I think everyone feels pretty much the same way.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top