Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

ever since I begun calculating thermodynamical cycles, I've had problems with determining the sign of the work along a particular bit of the cycle. Of course, I guess that an arbitrary cycle is 'bendy' and the sign of the work differential depends strongly on the coordinates, but usually the cycles I deal with consist of a couple of 'parts', for instance, adiabatics, isothermals, isochorics, etc. And whenever asked to calculate the total work done in a cycle (for instance to find the efficiency), I just kind of guessed the sign and managed to get by, but now when trying to understand this on a deeper level, this is coming back to haunt me.

So, suppose I have a thermodynamic cycle like this: LINK






How do I know the sign of the work along each of these paths?

share|cite|improve this question
I think it's work done on (or to) the system is positive and work done by the system is negative. I might have that backwards though – Jim Jun 25 '13 at 20:32
Related: and links therein. – Qmechanic Dec 24 '13 at 3:00
up vote 2 down vote accepted

General remarks.

Let $\delta W$ denote the differential work done by a system, so $\delta W$ is postive when the system does work on something else and negative when work is done by something else on the system. For a given process taking place over a path $\gamma$ in thermodynamic state space, the systematic way of determining whether work was done by or on the system is to determine the sign of $W$, the total work done by the system, which is given by $$ W = \int_\gamma\delta W $$ This can be computed in various ways depending on the system at hand, and the process it undergoes. The trick is to attempt to find an expression for $\delta W$ that allows for the efficient calculation of the integral for $W$.

Example - adiabatic compression.

Suppose,for example, that we want to determine the work done by the gas during process $1$ of your diagram. Recall that the first law of thermodynamics in differential form can be written as follows: $$ dE = \delta Q - \delta W $$ The sign convention here is that $\delta Q$ denotes the heat transferred to the system, and $\delta W$, again, denotes the work done by the system. Since process $1$ is adiabatic, we have $\delta Q = 0$ by definition. It follows that $$ W = -\int_\gamma dE = -\Delta_\gamma E $$ where $\Delta_\gamma E$ denotes the total change in energy along the path $\gamma$. Let process $1$ start at point $a$ and end at point $b$, then we can write this result as $$ W = -(E_b - E_a) = E_a-E_b $$ So to determine the sign of the work done, we simply need to know whether or not the internal energy of the gas increased (in which case $W<0$ so that work was done on the gas) or decreased (in which case $W>0$ so that work was done by the gas). How to we figure this out for this adiabatic process? Well take, for example, a monatomic ideal gas, and recall that for such a process, we have $$ T_aV_a^{\gamma-1} = T_bV_b^{\gamma-1}, \qquad \gamma = \frac{5}{3} $$ Then we see that since $V_b<V_a$, we have $T_b>T_a$; the temperature of the gas increased. But for a monatomic ideal gas, the internal energy can be written purely as a function of temperature and number of particles; $$ E = \frac{3}{2}NkT $$ so assuming the number of particles is fixed, the internal energy also increase, and therefore, $W<0$, so work was done on the gas.

share|cite|improve this answer
Thanks for the answer. BTW, when is it true that $\delta W=-pdV$? I recall this being a useful tool for quick calculations, but surely its applicability is limited? For example, in path (3), the definition $\Delta T = 0$ implies $\Delta E =0$, so $\delta Q = \delta W$. But I know of no way to integrate $\delta Q$, and since it's not a state variable, I can't just write $Q_b - Q_a$... – Spine Feast Jun 25 '13 at 21:02
@DepeHb Using the sign conventions I used in the response, one would have $\delta W = +PdV$. This can be used with ideal gases for example. And yea, for isothermal processes you would use that expression along with the ideal gas law to write $P$ as a function of $V$, and then you would explicitly perform the integral. – joshphysics Jun 25 '13 at 23:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.