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enter image description here

{Assume it's a Spherical Balloon}

The answer on the book states (7pV), but I'm definitely missing something. I've looked up online on what 'Workdone against atmosphere' means but there is little to zero answer on that.

It might be because I haven't understand fully what the question meant, but my assumption is:

If the diameter increases by twofold that means its volume increases by a scale factor of 8V. Which basically according to me, the answer should be 8pV, unfortunately the choice isn't there. Or Is this probably the same as saying the work done against friction in moving an object?

Any clarification would be highly appreciated.

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closed as off-topic by John Rennie, GiorgioP, ZeroTheHero, Jon Custer, Kyle Kanos Apr 22 at 17:42

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You are almost there. Just a small slip: The work done is $P\Delta V$. If the diameter doubles, the volume goes from $V$ to $8V$, so $\Delta V= (8-1)V=7V$. Thus $W= P 7V$.

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Imagine the atmosphere is just a really really big cylinder of gas. When I expand the balloon to 4 times the diameter how much does the volume of the balloon change ($V_{\mathrm{final}}-V_{\mathrm{initial}}$)? How much does the volume of the atmosphere change? If I change the volume of a gas by that much, at constant pressure, how much work did I do?

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