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A photon of visible light with energy 2 eV is absorbed by a macroscopic body held at room temperature. By what factor does $\Omega$ for the macroscopic body change?

My instincts tell me that I should be using the relations $S=kln\Omega$ and $TdS-PdV=dU$, but how can I use these to my advantage? Thank you!

Edit: as I commented, my lack of understanding is with regard to how I can calculate the number of initial and final microstates, as well as the assumptions we can make.

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  • $\begingroup$ $S = k Ln \Omega $ is entropy, as I am sure you know. When we include the $k $ factor, $S $ has units of energy divided by temperature, $\frac { J}{K}$. You have the thermodynamic identity and you have energy in electron volts, en.wikipedia.org/wiki/Electronvolt. $\endgroup$ – user108787 Aug 31 '16 at 15:48
  • $\begingroup$ @CountTo10 I understand that, but my inability to answer this question stems from not knowing how to use these equations. For instance, how can we figure out the initial and final number of microstates from just the arbitrary "macroscopic body"? What allows us to make assumptions about the differential change in volume? Thanks. $\endgroup$ – user107224 Sep 1 '16 at 2:11
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All you need is the statistical definition of temperature. \begin{align*} \frac{1}{T} := k\frac{\partial \ln \Omega(E)}{\partial E} \\ \frac{1}{T} = k\frac{1}{\Omega(E)}\frac{\partial \Omega(E)}{\partial E}\\ \\ \frac{\partial \Omega(E)}{\partial E} = \frac{\Omega(E)}{kT} \\ \Delta \Omega \approx \frac{\partial \Omega(E)}{\partial E}\Delta E \\ \end{align*}

All that is left is to substitute in for the two factors on the right hand side of the last line. You should find a proportionality constant.

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