# When an expansion of a system does work on its surroundings, is the sign for work negative?

This question comes to me from trying to explain the answer $1.67\ \mathrm{L}$ of this question:

A cylinder with a moving piston expands from an initial volume of $0.250\ \mathrm{L}$ against an external pressure of $2\ \mathrm{atm}$. The expansion does $288\ \mathrm{J}$ of work on the surroundings. What is the final volume?

• Work done on the surroundings is the force exerted on the surroundings dotted with the displacement of the surroundings. Are the force exerted on the surroundings and displacement of the surroundings in the same direction in this problem, or in opposite directions? – Chet Miller Nov 19 '16 at 13:00
• I can't tell. How does one tell that? – Danny Rodriguez Nov 19 '16 at 17:17
• The work being done is in the same direction , I would imagine. The piston only has in place to move. – Danny Rodriguez Nov 19 '16 at 17:25
• – Loong Nov 19 '16 at 18:27

It means the change in internal energy is negative, so you are on the right track. Because the system is doing work on the surroundings, that $288\ \mathrm{J}$ is a loss for the system and a gain for the surroundings.

One way to intuitively work your way through this is to think about what the signs on the terms must be in order for the problem to make sense.

You know the final volume is greater than the initial (it's an expansion of the system into the surroundings), so $\Delta\mathrm{V}$ has to be positive. You know that $w = -\mathrm{p_{ext}}\Delta\mathrm{V}$ is negative, because $\mathrm{p_{ext}}$ is positive. Thus, the sign of $\Delta\mathrm{E}$ is negative.

You want to solve

$$\Delta\mathrm{E} = -\mathrm{p_{ext}}\Delta\mathrm{V}$$

for $\Delta\mathrm{V}$ and add that to your initial volume of $0.25\ \mathrm{L}$.

Solving for $\Delta\mathrm{V}$:

$$-288\,\mathrm{J} = -2\ \mathrm{atm}\times\Delta\mathrm{V}$$

$$\Delta\mathrm{V} = 144\ {\mathrm{J\over atm}}$$

and then converting units

$$\Delta\mathrm{V} = \left(144\ {\mathrm{J\over atm}}\right)\cdot \left({\mathrm{L\cdot atm}\over 101.325\ \mathrm{J}}\right) = 1.42\ \mathrm{L}$$

we now solve for the final volume as (spoiler alert):

$$\mathrm{V_{final}} = 0.25\ \mathrm{L} + 1.42\ \mathrm{L} = 1.67\ \mathrm{L}$$