# Which form of the first law of thermodynamics should I use?

I usually wonder which thermodynamics first law is better to use ?

The one given by physics : $\Delta U=Q-\Delta W$

or the one by chemistry : $\Delta U=Q+\Delta W$

In other words, should I take the gas as my system and take every parameter in its terms?

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As a student it is the greatest doubt for me. And in exams (having all subjects like IIT-JEE simulators ) it is a big thing to see before marking an answer that the question belongs to physics or chemistry. –  ABC Apr 5 '13 at 10:43
Its not a bad question at all! I am an engineer and I never realized that chemists used a $\Delta U = \delta Q + \delta W$! I always use a $-\delta W$... –  drN Apr 5 '13 at 11:59
@drN: What if I told you that chemists look at dipoles the wrong way as well? :) –  Manishearth Apr 5 '13 at 12:22
That was also a confusion but now i have just got it . In chemistry the direction is towards the more electro-negative element, ie. -ve charge. :) –  ABC Apr 5 '13 at 12:39

There's no "better" or worse here. It's just that "work" in physics is defined differently than in chemistry.

In chemistry, all quantities follow this sign convention: They are positive if their effect is on the system. So, basically,

• $dU$ is (infinitesimal) energy imparted to the system by the surroundings
• $\delta Q$ is the heat passed to the system from the surroundings
• $\delta W$ is the work done on the system by the surroundings

In physics, the sign convention of $W$ is the opposite

• $dU$ is energy imparted to the system by the surroundings
• $\delta Q$ is the heat passed to the system from the surroundings
• $\delta W$ is the work done by the system on the surroundings

which means that $dU_C = \delta Q_C + \delta W_C$ ($C$ means chemistry) becomes $d U_P = \delta Q_P - \delta W_P$

Try to keep these conventions separate in your mind. Don't use the physics FLT for a chemistry problem and vice versa, many times problems specify values of $W,Q,U$ and expect you to know the sign convention.

Note that these are the IUPAC/IUPAP conventions. Some books (as @dmckee mentions, Feynman's Lectures is one of them) use different conventions. In such cases, just make note of the convention and remember that the FLT is just a statement of conservation of energy.

Here's the menemonic I used to remember it. It's not a great one, but it works:

Chemists are interested in supplying hear/energy/pressure to a reaction to make it occur. Thus, action done by the surroundings on the system is "good" or "positive.

Physicists are more interested in supplying heat/energy to a system and making it do work. So, supplying heat/energy is "good", and getting work out is "good".

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+1 for the lines after <hr>. :) –  ABC Apr 5 '13 at 11:55
Are you in IIT Bombay? –  ABC Apr 5 '13 at 12:00
@exploringnet: Yes, how did you guess? (Lets move this to Physics Chat, irrelevant here.) –  Manishearth Apr 5 '13 at 12:04
The Chemists/Physicists is not perfect. The Feynman Lectures takes $W$ to be the work done on the system. One must simply ask what conventions are in use and recall that this is a way of expressing the conservation of energy. –  dmckee Apr 5 '13 at 13:40
@dmckee: Right, edited, thanks :) –  Manishearth Apr 5 '13 at 13:43

Both the question and the accepted answer have the same terrible mistake: expression of the form $\Delta A$ does make sense only if $A$ is a state function.

IUPAC convention is that energy transferred to a system by any means is always considered to be positive, while energy removed from a system is considered to be negative. In accordance with this convention, when the first law is written as follows $$\Delta U = q + w \, , \tag 1$$ $q$ and $w$ stand for the heat supplied to the system and the work done on the system respectively. Both heat $q$ and work $w$ are thus by convention positive when energy enters the system, which results in an increase in internal energy, and negative when energy leaves the system, which results in a decrease in internal energy.

Now note that writing $\Delta$ in front of $q$ or $w$ is, as I already mentioned, a big mistake, and to explain why, we first look at an alternative form of the expression of the first law of thermodynamics.

For a quasistatic process in infinitesimals the first law takes the following form $$\mathrm{d} U = \text{đ} q + \text{đ} w \, . \tag 2$$ Note that infinitesimal heat and work transfers in a quasistatic process are denoted by $\text{đ}$ rather than $\mathrm{d}$, used to denote infinitesimal change in internal energy, to emphasize the fact that heat and work in contrast to internal energy are not state functions. Mathematically speaking, $\mathrm{d} U$ is an exact differential, while $\text{đ} q$ and $\text{đ} w$ are inexact differentials. As opposed to an exact differential, an inexact differential cannot be expressed as the differential of a function, i.e. while there exist a function $U$ such that $U = \int \mathrm{d} U$, there is no such functions for $\text{đ} q$ and $\text{đ} w$.

And the same is, of course, true for any state function $a$ and any path function $b$ respectively: an infinitesimal change in a state function is represented by an exact differential $\mathrm{d} a$ and there is a function $a$ such that $a = \int \mathrm{d} a$, while an infinitesimal change in a path function $b$ is represented by an inexact differential $\text{đ} b$ and there is no function $b$ such that $b = \int \text{đ} b$. Consequently, for a quasistatic process in which a system goes from state $1$ to state $2$ a change in a state function $a$, can be evaluated simply as $\int_{1}^{2} \mathrm{d} a = a_{2} - a_{1} = \Delta a$, while a change in a path function $b$, can not be evaluated in such a simple way, $\int_{1}^{2} \text{đ} b \neq b_{2} - b_{1} \neq \Delta b$. And thus, when integrating (2) we just write $q$ and $w$ as in (1) rather then $\Delta q$ and $\Delta w$.

For conventions see Quantities, Units and Symbols in Physical Chemistry (IUPAC Green Book) (PDF, 2.6 MB), Section 2.11 Chemical Thermodynamics, p. 56.

P.S. Inexact differentials are sometimes denoted using $δ$ instead of $đ$, but IUPAC recommends the later.

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