Download Benzene and toluene form nearly ideal solutions. At 300 K t

2007
Physical Chemistry 2
Homework assignment 3, solution
Problem 1:
Benzene and toluene form nearly ideal solutions. At 300 K the vapor pressure of toluene
is P∗ (toluene)=3.572 kPa and that of benzene is P∗ (benzene)=9.657 kPa.
(a) Compute the vapor pressure of a solution consisting of 0.6 mole fraction toluene and
0.4 mole fraction benzene.
(b) What is the mole fraction of toluene in the vapor above this liquid?
Solution:
(a) An ideal solution satisfies Raoults law for all components, i
Pi = xi Pi∗
So, the partial pressure of toluene in the vapor above the liquid solution will be Ptoluene =
0.6 · 3.572kP a = 2.143kP a and the partial pressure of benzene will be Pbenzene = 0.4 ·
9.657kP a = 3.863kP a. The total vapor pressure of the solution will, therefore, be P =
Ptoluene + Pbenzene = 2.143kP a + 3.863kP a = 6.006kP a.
(b) The mole fractions in the vapor can be obtained from the partial pressures. Since
benzene has higher vapor pressure, it will escape the liquid solution more readily than
toluene and therefore the vapor will have a higher mole fraction of benzene than the liquid
solution. In the vapor, xbenzene = 3.863/6.006 = 0.643 and xtoluene = 2.143/6.006 = 0.357.
This is the basis for distillation.
Problem 2:
Calculate the freezing point depression that occurs when table salt is added to a water/ice
mixture and the concentration of the salt in the solution becomes 1 M.
Solution
The freezing point depression is given by
∆T =
RT02
ln (1 − x)
∆Hf
where ∆T = Tf − T0
For water, ∆Hf = 6008J/mol at the freezing temperature of pure water, T0 = 273.15 K.
A solution that has the molar concentration of 1 M has mole fraction x = 1/55.5 = 0.018.
Plugging into the equation gives ∆T = −1.88 so the freezing temperature is lowered to
-1.88 ◦ C. You need a lot of salt to melt the ice on a cold day!
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Problem 3:
Consider a collection of N independent, distinguishable molecules that have two accessable
energy levels that differ in energy by ∆ǫ. The upper level is twofold degenerate. Take the
zero of the energy axis to be the lower energy level.
(a) Find the partition function for the molecules.
(b) Calculate the population of the two energy levels as a function of temperature and
discuss the two limits as the temperature approaches zero and infinity (explain the meaning
of your results).
(c) Calculate the average energy of the system as a function of temperature and discuss
the two limits as the temperature approaches zero and infinity (explain the meaning of
your results).
(d) Calculate the heat capacity as a function of temperature. Find the hight and low
temperature limits. Why are the high temperature results so different from the high
temperature results for translation, rotation and vibration (given by a constant larger
than zero)?
(e) Assume that the system has been prepared in such a way that the population of the
higher energy level is three times as large as the population of the lower energy state. While
Boltzmann statistics do not really apply in this case, use the Boltzmann distribution to
define an effective temperature for the system. Note and discuss the sign.
Solution:
∆ǫ
0
(a) The partition function for one molecule is
q =
3
X
i (states)
e
−βǫi
=
2
X
g(j) e−βǫj = 1 + 2e−β∆ǫ
j (levels)
The partition function for N distinguishable molecules is
Q = q N = (1 + 2e−β∆ǫ )N
(b) The population of level j, that is the number of molecules in level j, nj , is given by
the Boltzmann distribution as
g(j)e−βǫj
nj = N
q
so, for the ground state n1 = N/(1 + 2e−β∆ǫ ) and for the higher energy level n2 =
2N e−β∆ǫ /(1 + 2e−β∆ǫ ).
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The limit as the temperature approaches zero, T → 0, which corresponds to β → ∞,
becomes n1 → N/(1 + 0) = N , that is all the molecules will be in the ground state and
n2 → N ·0/(1+0) = 0. As the temperature becomes very large, T → ∞, which corresponds
to β → 0, the population of the ground state becomes n1 = N/(1 + 2) = N/3 and the
population of the higher energy level becomes n2 = N ·2/(1+2) = 2N/3. At infinitely high
temperature the molecules become equally distributed over the three states, irrespective of
the energy because the energy difference becomes insignificant with respect to the thermal
energy.
(c) The average energy of the system can be obtained in two different ways. One is to
differentiate the partition function
<E >= −
∂
∂
N ∂q
N
2N ∆ǫ e−β∆ǫ
−β∆ǫ
ln Q = −N
ln q = −
= −
(−2∆ǫe
)
=
∂β
∂β
q ∂β
1 + 2e−β∆ǫ
1 + 2e−β∆ǫ
The other method is to use the calculated population of the energy levels and sum over
population times energy
<E>=
X
nj ǫj = n1 ǫ1 + n2 ǫ2 = n1 · 0 +
j
2N ∆ǫ
2N e−β∆ǫ
∆ǫ = β∆ǫ
−β∆ǫ
1 + 2e
e
+ 2
which of course gives the same results.
The limit as the temperature approaches zero, is < E >→ 2N ∆ǫ · 0/(1 + 0) = 0 and
the limit as the temperature approaches infinity is < E >→ 2N ∆ǫ · 1/(1 + 2) = 2N ∆ǫ/3.
(d) The heat capacity is obtained by differentiating the energy
CV
∂<E>
1 ∂<E>
2N ∆ǫ ∂
=
= −
= −
2
∂T
kB T
∂β
kB T 2 ∂β
1
eβ∆ǫ + 2
=
2N kB (β∆ǫ)2 eβ∆ǫ
(eβ∆ǫ + 2)2
The limit as the temperature approaches zero, is CV → 2N kB (β∆ǫ)2 eβ∆ǫ /e2β∆ǫ →
2N kB (β∆ǫ)2 e−β∆ǫ → 0. Note that the exponential e−β∆ǫ goes to zero faster than β 2 goes
to infinity. The high temperature limit is CV → 2N kB (β∆ǫ)2 e−β∆ǫ → 0 · (1/(1 + 2)) = 0.
The reason the heat capacity becomes zero in the high temperature limit is that the system
saturates and there is no way of increasing the energy of the system as the temperature is
increased further. In the cases of translation, rotation and vibration there infinitely many
states and no upper limit on the energy of the levels.
(e) Given that n2 /n1 = 3. According to the Boltzmann distribution n2 /n1 = 2e−β∆ǫ .
Equating the two, gives 3 = 2e−β∆ǫ . After taking the logarithm of both sides, one gets
ln 3 = −∆ǫ/kB T , that is T = −∆ǫ/kB ln 3 which is a negative number. So, apparently the
temperature is negative! But, the system is really not in thermodynamic equilibrium and
the temperature is not defined.
Problem 4:
Go back to the experiment on heat capacity ratios and calculate
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(a) The vibrational contribution to the heat capacity of N2 molecules at room temperature
using the quantum statistical mechanical result for the partition function of a harmonic
oscillator (the vibrational temperature is 3393 K).
(b) Calculate the heat capacity ratio for nitrogen gas assuming the hight temperature
approximation (equipartition) is good enough for rotation and translation, but using the
quantum mechanical result from (a). How does the value you calculate compare with the
measured value you got in the lab? How important is the vibrational contribution to the
heat capacity ratio?
(c) Calculate the contribution of the four vibrational modes of CO2 to the heat capacity
of the gas at room temperature. The normal mode frequencies are: 1340 cm−1 for the
symmetric stretch, 2349 cm−1 for the asymmetric stretch and 667 cm−1 for the two bends.
(d) Repeat part (b) for CO2 gas.
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