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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 $-1.8 p_1 V_1$?

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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.

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