Change in accessible states relating to change in energy

Click For Summary
SUMMARY

The discussion centers on calculating the increase in the number of accessible states, denoted as Ω, when the internal energy of a system with 6 × 1024 degrees of freedom increases by 1%. The formula used is Ω = ENν/2, where ν represents the degrees of freedom. The user attempted to compute the ratio of Ω' to Ω using large exponent calculations but faced limitations with standard calculators. A suggestion was made to utilize logarithms to simplify the calculation process.

PREREQUISITES
  • Understanding of statistical mechanics concepts, particularly accessible states
  • Familiarity with the formula Ω = ENν/2
  • Basic knowledge of logarithmic functions and their properties
  • Experience with handling large numbers in calculations
NEXT STEPS
  • Learn how to apply logarithms to simplify complex exponentiation problems
  • Study statistical mechanics, focusing on the relationship between energy and accessible states
  • Explore computational tools for handling large numbers, such as Python's Decimal module
  • Investigate the implications of energy changes on thermodynamic systems
USEFUL FOR

Students and professionals in physics, particularly those studying statistical mechanics, thermodynamics, or anyone involved in calculations related to energy and accessible states in complex systems.

leroyjenkens
Messages
621
Reaction score
49

Homework Statement


A certain system has 6 × 10^24 degrees of freedom. Its internal energy
increases by 1%. By what factor does the number of accessible states increase?

Homework Equations


[tex]\Omega = E^{N\nu/2}[/tex]
[itex]\nu[/itex] is the degrees of freedom, and N is just 1, so we can ignore that. So the exponent is just 3x10^24
[itex]\Omega[/itex] is the number of accessible states.

The Attempt at a Solution


[/B]
First I replaced E with 10. And then E increased by 1% would be 10.1.
What I was going to try to do was divide 10.1^(6x10^24) by 10^(6x10^24) to get my answer. But no calculator in the world can do that.
So I used a calculator that can do big numbers (but not quite that big). I found that as I increased the 0's in the exponent, the exponent of my answer increased by some seemingly random number. I tried 10.1^(300)/10^(300) and then 10.1^(3000)/10^(3000), and then kept adding zeroes like that to see what kind of pattern I got.
What I got was, as I got up to 10 zeroes, was an answer of 3x10^(129641213). If I take away a zero from the exponents in the fraction, then I just lose an exponent in the answer. So if I had 24 zeroes in my exponents in my fraction, then the answer would be roughly 3x10^(1.29x10^22).

But this can't be the way to do this problem. Anyone have an alternative idea on how to solve this? Thanks.
 
Physics news on Phys.org
Try taking logarithms on both sides: the desired ratio is ##\Omega '/\Omega = \exp ...##
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
945
Replies
30
Views
5K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K