Vaindil
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 Mar 6 awarded Famous Question Jan 8 awarded Famous Question Jul 14 awarded Popular Question Feb 17 awarded Popular Question May 23 awarded Notable Question Apr 18 awarded Notable Question Oct 3 awarded Popular Question May 16 awarded Popular Question Apr 22 awarded Supporter Apr 22 comment Three polarizers, 45° apart Exactly what I was looking for. Thank you! Apr 22 accepted Three polarizers, 45° apart Apr 22 asked Three polarizers, 45° apart Oct 17 asked Adiabatic expansion Oct 10 awarded Student Oct 9 accepted Work Done by an Adiabatic Expansion Oct 9 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)! Oct 9 asked Work Done by an Adiabatic Expansion Oct 9 awarded Scholar Oct 9 accepted Work Done in an Isobaric Process Oct 9 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.