Heat capacity. The electronic heat capacity C e can be found by taking the derivative of Equation (19.18): For temperatures that are small compared with the Fermi temperature, we can neglect the second term in the expansion compared with the first and obtain.
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Thus the electronic specific heat capacity is 2.2 x 10-2 R. This small value explains why metals have a specific heat capacity of about 3R, the same as for other solids.
where the first term is the electronic contribution and the second is associated with the crystal lattice.
Figure 19.9 Sketch of the heat capacity of a metal as a function of temperature showing the electronic and lattice contributions.
S = 0 at T = 0, as it must be. The Helmholtz function F = U -TS is
N/V = 5.9 x1028 m-3 and TF= 65,000K .
P = 2/5 *5.9*1028 *(1.38*10-23) (6.5*104)
= 2.1*1010 Pa = 2.1*105 atm.
The temperature inside the core of a typical star is at the order of 107K.
The atoms are completely ionized at such a high T, which creates a hugh electron gas
The loss of gravitational energy balances with an increase in the kinetic energy of the electrons and ions, which prevent the collapse of star!
Example: The pressure of the electron gas in Sirius B can be calculated with the formula
Using the following numbers
Mass M = 2.09 × 1030 kg
Radius R= 5.57 × 106 m
Volume V= 7.23 × 1020 m3
Assuming that nuclear fusion has ceased after all the core hydrogen has been converted to helium!
there are electrons
i.e. is a valid assumption !
Thus: P can be calculated as
can be expressed as
To find the minimum U with respect to R
Using the following equation
(c) What is the average speed of the electrons in the fermion gas (see problem 19-4). Compare your answer with the speed of light.
1. H is a continuous function of the p.
2. If all the pi’s are equal, pi = 1/n; then H(1/n,…, 1/n) is a monotonic increasing function of n.
3. If the possible outcomes of a particular experiment depend on the possible outcomes of n subsidiary experiments, then H is the sum of the uncertainties of the subsidiary experiments.
one can expect that the function g( ) shall be a logarithm function.
H(1/2, 1/3, 1/6) = -K*[1/2ln(1/2) + 1/3*ln(1/3) + 1/6*ln(1/6)]
= -K*(-0.346 – 0.377 – 0.299) = 1.01K
from the decomposed procedure,
H(1/2,1/2) + 1/2H(2/3, 1/3) = -K*[1/2ln(1/2) + 1/2ln(1/2)]
-1/2*K*[2/3*ln(2/3) + 1/3ln(1/3)]
= -K(-0.346-0.346) – K/2*(-0.27 – 0.366)
H = - K*[p1ln(p1) + p2ln(p2)]
lim[u(x)/v(x)] as x approaches 0 equals lim[u’(x)/v’(x)]
lim[p1ln(p1)] = lim[(1/x)/(-1/x2)] = 0
differentiate equation - K*[p1ln(p1) + p2ln(p2)] against p1 and set the derivative equal 0
dH/dp1 = -K*[ln(p1) + p1/(p1) - ln(1- p1) – (1- p1)/ (1- p1)] = 0
which leads to p1 = 1/2
(a) English has the lowest entropy of any major language, and
(b) Shakespeare’s work has the lowest entropy of any author studied.