SebastianRM said:
Hey guys, so I am reading this book and on pages 89-90, the author says:
"Increasing temperature correspond to a decreasing slope on Entropy vs Energy graph", then a sample graph is provided, and both in that graph and in the numerical analysis given in page 87 the slope is observed to be an increasing slope too when the system increases in temperature. So I am somewhat confused.
Thank you in advance!
While the book is well-known, not everyone has it. So, it would have been good to include the graph you mention.
The horizontal axis running left-to-right [the lower axis] is the energy-count q_A in the A-EinsteinSolid (where the energy is U_A=q_Ahf).
The horizontal axis running
right-to-left [the
upper axis] is the energy-count q_B in the B-EinsteinSolid, where q_A+q_B=100 in this example.
As the energy-count q_A increases (to the right),
the entropy S_A is increasing (so the slope is positive)
but that
slope is decreasing [since increments are decreasing]
(so temperature T_A=\left(\frac{\partial S_A}{\partial U_A}\right)^{-1} is increasing).
As energy-count q_B increases (to the
left),
the entropy S_B is increasing (so the slope is positive)
but that
slope is decreasing [since increments are decreasing]
(so temperature T_B=\left(\frac{\partial S_B}{\partial U_B}\right)^{-1} is increasing).
Given an initial state (q_A,q_B), with q_A+q_B=100 for the combined system,
the system statistically evolves towards equilibrium (where S_{total} has slope zero).
Here equilibrium is at (q_A,q_B)=(60,40).
Since there are many more states near (60,40),
if we start at (30,70), the system statistically evolves to the right from (30,70), and
if we start at (70,30), the system statistically evolves to the left from (70,30).
