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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
up vote 1 down vote accepted

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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According to the international standard ISO 80000 Quantities and units – Part 5: Thermodynamics,

for a closed thermodynamic system

$$\Delta U = Q + W$$

where $Q$ is amount of heat transferred to the system and $W$ is work done on the system provided that no chemical reactions occur

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Terminology should be correct... All will be corrected itself

δq= Small heat given to the system

δW= small amount of work done on or by the system

ΔU = change in internal energy of the substance taking inside the system

And formula can be generalised for both chem n phy as

δq = δW + dU

Small Heat given to the system will be utilised in doing small amount of work and changing the Internal energy...

In physics if work is done by the system we'll take δW as positive..

In chem if work is done by the system we'll take δW as negative..

Trust me this formula will not mislead you....

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