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If the piston shown above moves upwards a distance 1 m the shaft work (according to the textbook solution to a related problem) is 1334 J. Shouldn't the sign of work be negative because the shaft moves upward (positive z direction) and the force is negative? According to the sign convention used by the textbook, work input is negative and work output is positive but I don't see how to determine the sign from this information.

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  • $\begingroup$ You are familiar with Newton's 3rd law of action-reaction, right? If the surroundings apply a force of 1334 N on the system, the system exert an equal and opposite force of 1334 N on the surroundings. And this force exerted on the surroundings is in the same direction as the piston displacement. So it is positive (shaft) work done on the surroundings. $\endgroup$
    – Chet Miller
    Commented Sep 12, 2020 at 12:46

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The sign for work depends on which version of the first law is used. For this version

$$\Delta U=Q-W$$

$W$ is positive when the gas does work (expands). For this version

$$\Delta U=Q+W$$

It is negative if the gas does work

The key is when the gas does work it reduces the internal energy. Both versions are consistent with that.

Your book appears to be using the first version of the first law.

Hope this helps

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