# Thermodynamic cycles, when is the work negative/positive?

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

Where:

(2)-isobaric

(3)-isothermal

(4)-isochoric

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

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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: physics.stackexchange.com/q/37904/2451 and links therein. – Qmechanic Dec 24 '13 at 3:00

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 $W$.
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.
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