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GR8677 #66
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Thermodynamics$\Rightarrow$}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.

Alternate Solutions
 ryanjsfx2014-09-21 13:08:27 I'm not sure my logic is sound but I got this just by process of elimination. We're looking for inverse of T, so eliminate A and B since they look normal (not inverted) expressions. If T changes, so does U [U = U(T)]. Eliminate C and D. That leaves E.Reply to this comment Hardik2008-11-06 13:11:39 Use statistical definition of temperature and entropyrnrnvoilaReply to this comment
socolenco_
2015-09-18 23:21:35
You have [$\\ ML^{2}$$\\ T^{-2}$$\\ K^{-1}$] for entropy and [$\\ ML^{2}$$\\ T^{-2}$] for internal energy and their ratio is$\\ K^{-1}$, which is exactly the reciprocal of the Kelvin temperature T°.\r\n
ryanjsfx
2014-09-21 13:08:27
I'm not sure my logic is sound but I got this just by process of elimination.

We're looking for inverse of T, so eliminate A and B since they look normal (not inverted) expressions.

If T changes, so does U [U = U(T)]. Eliminate C and D.

That leaves E.
spacemanERAU
2009-10-18 18:22:02
how do you know the volume is constant?
 neon372010-11-03 11:30:43 because $PdV$ needs to be zero in $dQ = dU + PdV$ So that we can use $dQ = dU$. Otherwise, the answer would be complicated and you would have to reciprocate the whole thing. And there is no complicated answer in the question.
 flyboy6212010-11-14 22:13:36 You don't know the volume is constant. What you know is that IF you hold the volume constant while differentiating the entropy with respect to internal energy, you get the inverse of the temperature. Whenever you take a partial derivative, you have to hold something constant.
Hardik
2008-11-06 13:11:39
Use statistical definition of temperature and entropyrnrnvoila
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, $T = \left(\frac{\partial U}{\partial S}\right)_{N,V}$.
 flyboy6212010-11-14 22:14:45 Perfect solution!
carlospardo
2007-10-03 13:10:49
All this is unnecessary. When obvious, select the answer with correct units.
 djh1012014-08-28 13:46:39 I can recall dU = TdS - PDV off the top of my head better than units of entropy (which I would probably derive from that equation anyway).
carlospardo
2007-10-03 13:10:48
All this is unnecessary. When obvious, select the answer with correct units.
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 $\frac{1}{T}= \left( \frac{\delta S}{\delta U} \right) _V$
jax
2005-12-03 10:59:32
You mean $(\frac{DS}{DU})_V$ is the correct answer, right?
 yosun2005-12-04 23:11:43 right. note that the above is given in terms of inverse Temperature --- a commonly used multiplier quantity in thermo.
 jax2005-12-05 06:51:18 Doesn't the question ask for inverse temperature? When I worked it out I got $\frac{1}{T} = (\frac{\partial S}{\partial U})_V$
 walczyk2011-03-10 21:54:37 at extremely low energies, a small change in energy amounts to a huge gain in entropy, so the inverse T is huge, and T is tiny.

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