Lnmm1 dmm t t3 slowly pouring sand

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Unformatted text preview: e sand is gone. I will show that this way we will return to the original state that has the temperature T1. Consider adding a small mass dm at the moment when the total mass of the sand and the piston is m. Using the same arguments as in the previous lecture, the change in temperature dT is given by T + dT = T or dT/T = (R/(R+ Integrating this, we get ln(T/T1) = (R/(R+ So if m=m1 (after we picked up all the sand) then T = T1. ))ln(m/m1) ))( dm/m) = T...
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This note was uploaded on 04/24/2011 for the course CH 52635 taught by Professor Makarov during the Spring '11 term at University of Texas.

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