GREPhysics.NET: GR8677 #66

GR8677 #66

Problem

GREPhysics.NET Official Solution

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.

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Comments

carlospardo
2007-10-03 13:10:49

All this is unnecessary. When obvious, select the answer with correct units.

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carlospardo
2007-10-03 13:10:48

All this is unnecessary. When obvious, select the answer with correct units.

Reply to this comment

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 $\frac{1}{T}= \left( \frac{\delta S}{\delta U} \right) _V$

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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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