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A system was given a small amount of thermal energy dE, and its number of states G grew by 25%. How can I find the system temperature?

The system contains gas particles, I know that $dE << U$

I thought about using the entropy $\sigma$ = $ln(g)$ when $g$ is the number of states

$\frac{1}{T} = K_b \frac{d\sigma}{dU}$

when $dU = dE $ and the entropy $\sigma$ is $ ln(g)$ so $d\sigma = ln(1.25g)-ln(1g) = ln(1.25)$

we get - $ T = \frac{dE}{K_b\cdot ln(1.25)}$

and the Bolztmann constant units are $J/K$ so it turns out okay units wise, but apparently its not the correct answer. Have I messed up something?

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Why do you think this is incorrect? –  Yrogirg Apr 19 '13 at 13:13
    
the correct answer should be 2636.53K –  YNWA Apr 19 '13 at 14:09

1 Answer 1

Here is the correct solution $\frac{1}{T}=K_b \cdot \frac{d\sigma}{dU}$

now we take $\sigma=ln(g)$ -> $d\sigma=\frac{dg}{g}$ and that it $25$% . substituting this into the equation yields the correct answer. It makes sense, but still I dont get why is that "more" correct then my first way?

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$0.223 \ldots = \ln(1.25g)-\ln g \approx \frac{d \ln g}{d g} \Delta g = 0.25$ They are practically the same and both are approximations. None is more correct. –  Yrogirg Apr 20 '13 at 20:37

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