Question

A system with 20,000 molecules has two energy levels (A and B). At first, energy level A is populated by 60% of the molecules. After a reversible process, energy level A is populated by 90% of the molecules and the system is at 373 K.

a. What is the difference in energy between the two levels?
Provide your answer in J/mol.

b. What is the change in the total energy of the system?

c. What is the change in entropy?

Hint: After the reversible process, the system should reach
equilibrium and follows the Boltzmann distribution.

Answer #1

Suppose that a molecle has just two energy levels. The ground
state is singly degenerate and the excited state is triply
degerate. If the enrgy difference between the two states (on a
molar basis) is 52 kJ/mol at what temperature will the two states
be equally populated?
QUANTUM CHEMISTRY

Consider an isolated system containing two blocks of copper with
equal mass. One block is initially at 0oC while the other is at
100oC.They are brought into contact with each other and allowed to
thermally equilibrate. What is the entropy change for the system
during this process? The heat capacity for copper is Cp =
24.5 [J/mol k]

A system contains two energy levels separated by 300 cm–1 . (a)
What are the relative populations of the upper and lower levels if
the temperature is 1 K? What if the temperature is 1×105 K? (b)
What is the temperature of the system when the population of the
upper energy state is half that of the lower energy state?

1. A molecule has a ground state and two excited electronic
energy levels all of which are not degenerate. The energies of the
three states are E = 0, E1 = 1x10^-20 J and E2 = 2x10^-20 J.
Calculate the partition functions at 298 and 1000K. What fraction
of the molecules is in each of the three states at these
temperatures?

Diamond
a. At 298 K, what is the Gibbs free energy change G for the
following reaction? Cgraphite -> Cdiamond
b. Is the diamond thermodynamically stable relative to graphite
at 298 K?
c. What is the change of Gibbs free energy of diamond when it is
compressed isothermally from 1 atm to 1000 atm at 298 K?
d. Assuming that graphite and diamond are incompressible,
calculate the pressure at which the two exist in equilibrium at 298
K.
e....

A mole of two-state systems each has energy difference 2.346 ×
10-21 J.
1)
At what temperature are 90% of the two-state parts in their
low-energy states?
T90%=
K
2)
What is the heat capacity at T= 17 K?
CV=
J/K
3)
What is the heat capacity at T= 85 K?
CV=
J/K
4)
What is the heat capacity at T= 170 K?
CV=
J/K

± Free Energy and Chemical Equilibrium
The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG, using the
following equation:
ΔG∘=−RTlnK
where T is standard temperature in kelvins and
R is equal to 8.314 J/(K⋅mol).
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the...

The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG∘, using the
following equation:
ΔG∘=−RTlnK
where T is a specified temperature in kelvins (usually
298 K) and R is equal to 8.314 J/(K⋅mol).
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the
equilibrium values....

The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG∘, using the
following equation:
ΔG∘=−RTlnK
where T is a specified temperature in kelvins (usually
298 K) and R is equal to 8.314 J/(K⋅mol).
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the
equilibrium values....

In class we discussed a reversible, isothermal compression of an
ideal gas. The initial point (1) is an ideal gas in equilibrium at
pressure 1 (P1), volume 1 (V1) and temperature (T), and the final
point (2) is an ideal gas in equilibrium at pressure 2 (P2), volume
2 (V2) and temperature (T), where V2
d) For each differential step, the change in entropy is given by
dS=qrev/T =!!!"#! . Since T is constant , this express ion can be...

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