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  GR8677 #66
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GREPhysics.NET Official Solution    Alternate Solutions
Verbatim question for GR8677 #66
Thermodynamics}First Law

Recall that dQ=TdS=PdV+dU, where work done by the system is positive and heat input into the system is positive. For constant volume, the equation becomes dQ=TdS=dU.

\Rightarrow T=\left(\frac{\partial U}{\partial S}\right)_V. The reciprocol gives the right answer.
See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
Hardik
2008-11-06 13:11:39		Use statistical definition of temperature and entropyrnrnvoilaAlternate Solution - Unverified
Comments
Hardik
2008-11-06 13:11:39		Use statistical definition of temperature and entropyrnrnvoilaAlternate Solution - Unverified
student2008
2008-10-12 06:31:38		Actually, one can recall the general differential expression for the energy of a system dU = TdS - PdV + \mu dN. Thus, T = \left(\frac{\partial U}{\partial S}\right)_{N,V}.NEC
carlospardo
2007-10-03 13:10:49		All this is unnecessary. When obvious, select the answer with correct units.NEC
carlospardo
2007-10-03 13:10:48		All this is unnecessary. When obvious, select the answer with correct units.NEC
WTarrasque
2006-11-03 19:12:23		In the compiled solutions PDF, the reciprocal is only taken of one side. Thus it has:

\frac{1}{T}=\left( \frac{\delta U}{\delta S} \right) _V instead of \frac{1}{T}= \left( \frac{\delta S}{\delta U} \right) _V 		NEC
jax
2005-12-03 10:59:32		You mean (\frac{DS}{DU})_V is the correct answer, right?
yosun
2005-12-04 23:11:43		 right. note that the above is given in terms of inverse Temperature --- a commonly used multiplier quantity in thermo.
jax
2005-12-05 06:51:18		 Doesn't the question ask for inverse temperature? When I worked it out I got \frac{1}{T} = (\frac{\partial S}{\partial U})_V
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