# Steven H

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 Oct3 awarded Popular Question May16 awarded Popular Question Apr22 awarded Supporter Apr22 comment Three polarizers, 45° apart Exactly what I was looking for. Thank you! Apr22 accepted Three polarizers, 45° apart Apr22 asked Three polarizers, 45° apart Oct17 asked Adiabatic expansion Oct10 awarded Student Oct9 accepted Work Done by an Adiabatic Expansion Oct9 comment Work Done by an Adiabatic Expansion If I use $pV = RT$, then $p = \frac{RT}{V}$. Using $\delta w = pdV$, then $w = RT \int{\frac{1}{V}} dV$. Solving this gives me $w = RT\ln(5)$, which, after plugging numbers in, gives me $w = (287)(293)\ln5 = 1.35339 * 10^5$. And I was just sent an email saying that the book's answer is wrong and this one is correct. Thank you so much for your help, I really appreciate it (and I do completely get it now)! Oct9 asked Work Done by an Adiabatic Expansion Oct9 awarded Scholar Oct9 accepted Work Done in an Isobaric Process Oct9 comment Work Done in an Isobaric Process Wait -- is it really as simple as $(RT)/P=V$ (which gives me the change since one is 0°)? It would be divided by $p$ and then again multiplied by that, so they cancel out. This actually gives what the book says is the right answer. Oct9 awarded Editor Oct9 revised Work Done in an Isobaric Process deleted 14 characters in body Oct9 asked Work Done in an Isobaric Process