(a) Give an expression for the chemical potential mu of an ideal classical monat
ID: 1405355 • Letter: #
Question
(a) Give an expression for the chemical potential mu of an ideal classical monatomic gas at temperature T and pressure P. (See notes Section 5c). Show that the chemical potential of Argon gas (atomic mass 40) at 25 Celsius (NOT Kelvin) and pressure 1 atmosphere (= 10^5 pascal) is -0.414 eV. (b) A monatomic gas, when in contact with a solid surface, may deposit atoms onto the surface (adsorption). Assume that adsorption occurs at well-defined sites on the surface. Each site can adsorb an atom in two configurations. In configuration # 1 the atom is bound to the surface with an energy epsilon 1 = -0.35 eV and in configuration #2 it is bound with energy epsilon 2 = -0.30 eV. It is not possible for more than one atom to bind at a given site. (i) What are the possible states (N, i) and energies E i^(N) of a site? (Do not neglect the important case of an unoccupied site!) (ii) In terms of T and the chemical potential mu give an expression for the probability that a site is occupied by an atom in configuration #2. (iii) Find the probability in (ii) above when the surface is in contact with Argon gas at T = 25 Celsius and pressure P = 1 atmosphere (= 10^5 Pascal). (You can regard the gas as a heat bath determining mu and T for the site). If there are many sites on the surface, what percentage of them are occupied in configuration # 2? (iv) Under the same conditions as in (iii), what percentage of the sites are empty (unoccupied by an Argon atom)?Explanation / Answer
a)µ= µ* + RT ln P
µ= Chemical Potential
µ*=Standard Chemical Potential
R=Gas constant
T=Temperature
P=Pressure
µ=-KT ln[(2mkt/h^2)^(3/2) * KT/P]-------------------------------(1)
Standard Values of the given Variables:
R=8.314 Joule/Mole Kelvin
K=1.38*10^(-23)Joule/Kelvin
H=6.62*10^(-34)Joule Sec
M=40*1.66*10^(-27)kg=6.63*10^(-26)kg
T=298.15k
P=10^5 Pa
Substitute all the values in equation (1) we get
µ= -0.414 ev
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.