Note: the "electron volt" = eV is a commonly used unit of energy
in atomic physics. 1 eV = 1.6022 × 10^(-19) J
Problem A: Consider the molecules HCl and I2 (diatomic iodine).
In HCl the energy difference between the ground state
and the vibrationally excited state is: .37 eV
In I2 the energy difference between the ground state
and the vibrationally excited state is: 0.027 eV
At 25 C, find the ratio of the number of molecules in the excited state
to the number of molecules in the ground state for HCl.
Do the same for I2.
Which gas is more likely to have f=3+2+2 at this temperature?
Problem B: seeking hardcopy of computer-drawn plots or *accurate*
sketches of calculator-drawn plots. Make sure the units (J on y-axis,
K on x-axis for the first plot; J/K on the y-axis for the second)
and values on the axes are clearly denoted.
Consider a molecule with exactly two states with energy
E0=0 eV and E1=.01 eV.
Plot the internal energy of a mole of such molecules for
temperatures from 10 to 1000 K (if possible use a log scale for T).
The specific heat at constant volume is just the derivative
with respect to temperature of the internal energy calculated above.
Plot the specific heat of the one mole from 10 to 1000 K.
(If you can't do a log scale, only plot 10 to 500 K)
Problem C: Using Mathematica to compute thermodynamic functions.
NoteA: use the letters (symbols) for numerical quantities
and only in the end assign the actual numerical values.
NoteB: mostly you should be able to copy & paste these commands
into Mathematica!
Consider our simple harmoic oscillator energy levels (with
zero-point energy removed):
energy= e n
Where e is some small constant, and n is a whole number: n=0,1,2...
In order to calculate actual probabilites (rather than ratios)
we need to find the sum over states (partition function):
Sum[Exp[-e n/(k T)],{n,0,Infinity}]
which Mathematica can do.
In order to then calculate the average energy, we need to find
the sum of probability times energy (see Eq. T7.25):
avgE=Sum[e n Exp[-e n/(k T)],{n,0,Infinity}]/Sum[Exp[-e n/(k T)],{n,0,Infinity}]
which Mathematica can do.
In order to find the specific heat, we need to find the
change in energy (i.e., derivative) with temperature:
c=D[avgE,T]
which Mathematica can do.
Consider a mole of such oscillators:
na = 6.0221 10^23
with energy spacing .01eV:
e=.01 1.6022 10^-19
with Boltzmann constant:
k=1.3807 10^-23
Plot the energy and specific heat for this material
for a range of temperatures: from 0 to 300...something like:
Plot[na avgE ,{T,0,300}]
Turn in hardcopy of these two plots. Compare the plot
of avgE to Figure T7.9...what is different?