The work done during a reversible compression - if $V_\mathrm{i} = V_1$ (initial) and $V_\mathrm{f} = \frac{1}{6}V_1$ (final) - of an ideal gas is?

I chose $1.8 p_1 V_1$. Is it right or should it be $-1.8 p_1 V_1$?


1 Answer 1


PV work is defined as $$w = -p\Delta V,$$ where $p$ is pressure and $\Delta V$ is $V_\mathrm{final} - V_\mathrm{initial}$.

It has this form because in chemistry, work is defined from the perspective of the system - if it is compressed, we say that we did work on the system and so the sign of work is positive.

In this example, $V_\mathrm{f}$ is less than $V_\mathrm{i}$, so $\Delta V < 0$. If $\Delta V$ is negative, then $w$ must be positive.

This means your answer should have a positive sign. However, I get a different number from $1.8$ - you should double-check that.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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