# Question about first law of thermodynamics [duplicate]

Right now I am learning about the first law of themodynamics. But there is a problem which bother me a bit.

As in my chemistry lesson, the teacher introduces $$\Delta E =$$ $$Q+W %$$.

However in the physics textbook, it introduces $$\Delta E =$$ $$Q-W %$$

Is the difference between the both equation is due to different perspectives? Like for the first one is the work done on the system by surrounding while the second one is work done of the system on surrounding?

• Does this answer your question? Work term in First Law of Thermodynamics Also on Chemistry SE: chemistry.stackexchange.com/q/15571 Apr 12 at 16:31
• @Rishi The linked post asks a different question, "Is the work done by system on surroundings is always equal to work by the surroundings on the system?" This is not the same as to why some books say $E=Q+W$ but others $E=Q-W$. Apr 12 at 18:03
• @Themis Probably an even older duplicate: physics.stackexchange.com/questions/60117/… Apr 13 at 5:31
• @GiorgioP-DoomsdayClockIsAt-90, yes, the post you have linked is about the same issue. Apr 13 at 12:06

The sign convention for work has been a problem that adds to the confusion about thermodynamics.

Work, like heat, is energy in transit from one system to another, and we use a sign convention to indicate this direction, with the emphasis on convention. Most current engineering textbooks use the convention that work that goes in the system is positive. This is the opposite of the older convention, still used in some chemistry and physics books, where the work is taken to be positive if it goes from the system to the surroundings. The older convention is motivated by the situation in heat engines: we put heat into the system (which we count as positive) and receive work out of it (which we also count as positive. In this convention heat and work have opposite signs when they are transferred in the same direction and the same sign when they are transferred in opposite directions.

In the modern convention, which I personally consider to be less prone to confusion, all energy that goes in the system is positive and all energy that comes out is negative, regardless of what is the form of that energy.

That both conventions are still in use means that you must be careful to understand which convention your source is working with. Once work and heat have been replaced by thermodynamic functions ($$dW =- P dV$$ in the new convention, $$dW=+PdV$$ in the old one, $$dQ=T dS$$ in both conventions), the resulting equation $$dE = T dS - P dV$$ is independent of sign conventions.

Your view seems correct. Pay attention to how the contribution of the work $$W$$ is treated when a concrete problem is addressed.

• Yes. There are different conventions in Physics and Chemistry. See en.wikipedia.org/wiki/… Apr 12 at 16:29
• @mike stone,Oh my god that's extremly helpful. Since I use a Cantonese version of Wiki, it does not provide me there is a sign convention of the equation. After reading it, it prove my approach is correct!!! Thx for the help !!
– YH W
Apr 12 at 16:51

These formulas are intended to explain the relationship between work, heat and energy. Simply memorizing the formulas is not going to help you. Think of it in terms of work done on a system and work done by a system.

The first formula says the total energy of a system is the net heat exchanged from the system minus the thermodynamic work done by the system.

The second formula says the total energy of a system is the net heat exchanged from the system plus the work done on the system.

I have never liked the textbook definition of the first law of thermodynamics, as some textbooks define work leaving a system as positive while some textbooks define work leaving a system as negative. This leads to confusion regarding the sign to assign to the work term.

The first law of thermodynamics is a version of the law of conservation of energy. Assuming that all energy terms are positive, this leads to a formulation that leaves less room for confusion. That formulation is:

$$Q_{in} - Q_{out} + W_{in} - W_{out} = \Delta E$$

This formulation is just as valid as the formulation typically found in textbooks, and it decreases the chance of error when dealing with devices such as heat pumps, where there is $$Q_{in}$$ AND $$W_{in}$$, as well as $$Q_{out}$$.

• It is not always simple to distinguish between work that goes in and work that goes out. We can have paths on which the works flips sign. By your method we would have to break the path into segments of strictly positive or negative work, which is I think an unnecessary step. By the way, when you say $W_{in}-W_{out}$ you are adopting the modern convention that work (and heat) that goes in is positive, so at the end we are both using the same convention. Apr 12 at 18:11
• @themis, during my industrial experience working on chemical engineering problems, it was always easy to determine if work was entering the process or leaving the process. Do you have any real world examples? Apr 13 at 2:12
• Not an industrial problem, but if you wanted to calculate the work on a path where volume changes, say a a $sin$ function of time, wouldn't you have to break the path into segments monotonic in volume so that work is either positive or negative? Obviously a contrived example but I'm just trying to make the point. As a secondary point, how do you go from $Q_{in} - Q_{out} + W_{in} - W_{out} = \Delta E$ to $dE = T dS - P dV$? Apr 13 at 5:14