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I know that the equation for the work done on a gas is:

$W=-\int pdV$

So then if you compress the gas, dV is negative and thus the work on the gas is positive, vice versa when it expands.

But this literally makes no sense to me in this problem:

"A 0.90 kg block of aluminum is heated at atmospheric pressure so that its temperature increases from 22.0°C to 40.0°C."

Pressure is constant, so it ends up being $W = -p\triangle V$. Plug in $\triangle V=3\alpha V_0 \triangle T$ and solve (given density and coefficient of volumetric expansion or whatever),

The work shows that it's negative (though I'm not actually sure why I'd even use that equation but that's what the solution says to do), but that makes no sense to me at all. How come? From my perspective you're literally giving heat to the block of aluminum.

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The sign convention you are using is that the work is positive when it is work done on the gas and negative when it's work done by the gas. When you compress a gas you do work on it i.e. the energy involved in the work flows from you into the gas. Likewise if you allow the gas to expand it does work on you i.e. the energy involved in the work flows from the gas into you.

In this case the aluminium block is expanding so it is doing work on you, and using your sign convention the work is negative.

It is certainly true that you are adding heat to the case, but this is not the same as doing work. In this case the increase in internal energy of the aluminium will be less than the amount of heat added because some of that heat flows out of the aluminium as the work the aluminium does on you.

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