Change in number of microstates

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SUMMARY

The discussion focuses on calculating the relative increase in microstates, represented as ##\frac{\Delta \Omega}{\Omega}##, for an isolated macroscopic system at 300K that absorbs a photon with a wavelength of ##\lambda = 550nm##. The energy of the photon is determined using the equation ##E = \frac{hc}{\lambda}##. The relationship between entropy and microstates is established through the equation ##S = k \ln(\Omega)##, leading to the differential ##d \ln(\Omega) = \frac{hc}{\lambda T}##. The key takeaway is that the differential of ##\ln \Omega## suffices to find the relative increase without needing to calculate ##\Delta \Omega## directly.

PREREQUISITES
  • Understanding of thermodynamic concepts, specifically entropy and microstates.
  • Familiarity with the equation for photon energy, ##E = \frac{hc}{\lambda}##.
  • Knowledge of the relationship between entropy and microstates, ##S = k \ln(\Omega)##.
  • Basic calculus, particularly differentiation of functions.
NEXT STEPS
  • Study the concept of differentials in calculus, focusing on their application in thermodynamics.
  • Explore the implications of the Boltzmann entropy formula, ##S = k \ln(\Omega)##, in statistical mechanics.
  • Investigate the relationship between temperature, energy, and microstates in macroscopic systems.
  • Learn about the physical significance of microstates and their role in determining thermodynamic properties.
USEFUL FOR

Students and professionals in physics, particularly those studying thermodynamics and statistical mechanics, as well as anyone interested in the relationship between energy absorption and microstate changes in isolated systems.

Kara386
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Homework Statement


An isolated macroscopic system at 300K absorbs a photon with ##\lambda = 550nm##. What is the relative increase ##\frac{\Delta \Omega}{\Omega}## in microstates.

Homework Equations

The Attempt at a Solution


The energy of the photon is ##E = \frac{hc}{\lambda}## so that would be the change in internal energy. And ##dU = TdS - pdV## but I assume ##dV = 0## so ##dU = TdS##.

##S = k \ln(\Omega)## so I think ##dS = k d\ln(\Omega)##:

##d \ln(\Omega) = \frac{hc}{\lambda T}##
If that's all ok, I don't know how to get from there to ##\Delta \Omega## or how to get the relative increase in ##\Omega##! Thanks for any help!
 
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Kara386 said:
##d \ln(\Omega) = \frac{hc}{\lambda T}##
...I don't know how to get from there to ##\Delta \Omega## or how to get the relative increase in ##\Omega##! Thanks for any help!
You just need to carry out the differential of ##\ln \Omega##. You don't need to find ##\Delta \Omega##, you only need ##\frac{\Delta \Omega}{\Omega}##.
https://en.wikipedia.org/wiki/Differential_of_a_function
 
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TSny said:
You just need to carry out the differential of ##\ln \Omega##. You don't need to find ##\Delta \Omega##, you only need ##\frac{\Delta \Omega}{\Omega}##.
https://en.wikipedia.org/wiki/Differential_of_a_function
Thank you for that link! Been trying to find out what it's called when you put a 'd' in front of variables for ages. And how to then actually do something with it. Thanks for your help!
 

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