NAME:
KEY
Section Number: 10
CHEMISTRY 418, Fall, 2011(11F)
Midterm Examination #1, October 6, 2011
Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILY
understand your work; only work on the exam sheet will be considered. Write answers, where appropriate, with reasonable numbers of
significant figures. You may use only the handbook of “Essential Data and Equations for a Course in Physical Chemistry”, a calculator, and
a straight edge.
1. (20 points)
DO NOT WRITE
a) (10 points). Calculate the pressure of 0.635 g of copper in a gas phase, in a closed 10L container at IN THIS SPACE
3000°C (above copper boiling point) according to the ideal gas law.
p. 1________/20
Answer:
p. 2________/20
For any ideal gas: PV = nRT
n(Cu ) =
p. 3________/20
0.63 g
0.635 g
=
= 0.01 mol
M ( Ar ) 63.5 g / mol
p. 4________/20
p. 5________/20
At 3000°C:
P(Cu ) =
=============
n(Cu ) RT 0.01 mol × 8.3144 J / mol / K × 3273.15 K
J
=
= 27214.3 3 = 27214.3 Pa
3
V
m
m
10 L × 0.001
L
p. 6_______/5
(Extra credit)
=============
TOTAL PTS
= 0.27 atm
/100
b) (10 points). It is not easy to find van der Waals constants for copper gas. However, it is often
sufficient to include the repulsive interactions only into the gas law to improve the description significantly. Use the
following equation of state to determine the pressure exerted by 0.635 g of copper in a gas phase, in a closed 10L
container at 3000°C (above copper boiling point), same system as above:
P=
RT
, all the constants and variables have their usual meanings and b is excluded volume (volume occupied by the
Vm − b
molecules of gas) that can be calculated as Vmolecule =
4 3
4
π r or Vmol = π r 3 × N 0 , based on the van der Waals radius of
3
3
copper atom of 1.4Å (N0 is Avogadro’s number).
P=
RT
=
Vm − b
8.3144
J
× 3273.15 K
mol × K
(
)
3
4
m3
molcules
10 L × 0.001
/ 0.01mol − π × 1.4 × 10 −10 m × 6.022 × 10 23
3
L
mol
=
J
× 3273.15K
J
mol × K
= 27214.5 3 = 27214.5 Pa The number that is very slightly different from the ideal gas
4
m
1mol − π × 16.5 × 10 −7 mol
3
8.3144
law approximation. Of course, it is misleading because at these conditions, attractive forces will definitely play a role and also
the van der Waals radius that I provided is for solid copper, not for a gas phase atom
NAME:
CHEM 443, Midterm Exam #1, Fall, 2005, page 2
2. (20 points) For each expression given in the left column, chose the appropriate description from the right column
1) ΔU = q + w
__b__
⎛ G⎞
⎜∂ ⎟
H
2) ⎜ T ⎟ = − 2
T
⎜ ∂T ⎟
⎜
⎟
⎝
⎠P
__ m__
3)
TdS ≥ dq
4)
η=
__k__
Thot
Thot − Tcold
__f__
⎛ ∂T ⎞
⎛ ∂P ⎞
⎟ = −⎜ ⎟
⎝ ∂V ⎠ S
⎝ ∂S ⎠V
_i_
5) ⎜
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
m)
n)
o)
p)
q)
r)
Collisional sectional
1st law of thermodynamics
3rd law of thermodynamics
The heat capacity differential
The law of quantum entanglement
Coefficient of performance of a heat
pump
The Claus inequality
The Poisseil’s law
An example of Maxwell relation
Mean ionic activity coefficient
The Clausius inequality
An example of Machiavellian relation
Gibbs-Helmholtz equation
Stokes-Einstein equation
Daniell equation
Gibbs-Smolukowski equation
The Clausius uncertainty
Coefficient of sale of an Otto cycle engine
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NAME:
CHEM 443, Midterm Exam #1, Fall, 2005, page 3
3. (20 points) Strontium carbonate is a salt that is commonly used in many applications. One of them is in the fireworks (it
Θ
gives them red color). This is the reaction that could be used to determine the ΔH f for this salt:
3
Sr ( s ) + C ( graphite, s ) + O2 ( g ) → SrCO3 ( s ) ;
2
Θ
however, most common methods would be really difficult to use to determine ΔH f (SrCO3 ( s ) ) directly. Use Hess’ Law and
the following set of reactions (with their respective enthalties of reactions) to determine ΔH f (SrCO3 ( s ) )
Θ
1)
1
Sr ( s ) + O2 ( g ) → SrO ( s ), ΔH I = −592 kJ
2
2)
SrO( s ) + CO2 ( g ) → SrCO3 ( s ), ΔH II = −234 kJ
3)
C ( graphite, s ) + O2 ( g ) → CO2 ( g ), ΔH III = −394 kJ
Answer:
If you simply add all three reactions, you will get the reaction of strontium carbonate formation. Thus:
ΔH Θf (SrCO3 ( s ) ) = ΔH I + ΔH II + ΔH III = −592kJ − 234kJ − 394kJ = −1220kJ
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NAME:
CHEM 443, Midterm Exam #1, Fall, 2005, page 4
4. (20 points). Calculate the final temperature following an adiabatic reversible compression of 2 moles of monoatomic ideal
gas with a heat capacity of CV = 1.5R from 100 L and 300 K to 10 L.
CVm
dV
dT = − R m (page 5-6)
T
Vm
T2
CVm
V
dT = − R ln 2
T
V1
T1
∫
T2
or
CV
V
dT = −nR ln 2
T
V1
T1
∫
CVm ln
⎛ T2
⎜⎜
⎝ T1
⎛V
T2 = T1 × ⎜⎜ 2
⎝ V1
⎞
⎟⎟
⎠
−
R
CVm
J
mol × K
J
3
×8.3145
mol × K
2
T2
V
= − R ln 2
T1
V1
⎞ ⎛ V2
⎟⎟ = ⎜⎜
⎠ ⎝ V1
⎞
⎟⎟
⎠
−
R
CVm
8.3145
⎛ 10 L ⎞
= 300 K × ⎜
⎟
⎝ 100 L ⎠
−
= 300 K × (0.1)
−2
3
300 K
=
3
(0.1)2
= 1392.5 K
This example used 1.5R as a molar heat capacity at a constant volume. However, because of the way the problem was
stated, anyone who used it as a heat capacity of two moles also received a full credit.
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NAME:
CHEM 443, Midterm Exam #1, Fall, 2005, page 5
5. (20 points). Calculate absolute molar entropy of solid potassium chloride (KCl) at 400°C and 1 bar. Use any
information available in the Blue Book and assume that the heat capacity of KCl can be approximated as a constant within
the temperature interval studied.
At constant pressure:
S
Θ
m , 673.15 K
=S
Θ
m , 298.15 K
773.15 K
+
∫
298.15 K
C p ,m ( KCl , s )
T
dT = 82.6
673.15 K
J
J
J
ln
+ 51.3
= 128.9
mol × K
mol × K 298.15 K
mol × K
This is very substantial. The number is positive, since it is absolute entropy. It is also higher than the entropy of the same
solid at room temperature, as the entropy should increase with temperature.
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NAME:
CHEM 443, Midterm Exam #1, Fall, 2005, page 6
6. (5 points, extra credit) Imagine a system of four compartments of equal volume V connected by valves (for example
four identical cylinders connected to a manifold). Each compartment contains 1 mole of a different gas at the same
pressure of 1 atm. This is an experiment we will do: open the valve between 1st and 2nd cylinder very slowly and wait till
the equilibrium is established between gases A and B. Than open the valve connecting the resulting reservoir to the 3rd
cylinder. After the equilibrium is established, open the 4th valve and wait for the entire system to come to equilibrium. You
can assume that the behavior of all gases can be described by the ideal gas law and that the temperature of the entire
process is 35°C. Calculate the entropy of mixing.
Answer: Entropy is a state function so all we need is to calculate the difference between the initial and final states.
According to equation on page 5-3 of your handbook,
ΔS mix = − R(n A ln X A + n B ln X B ) for two gases. Adapting this equation for four gases:
ΔS mix = − R(n A ln X A + n B ln X B + nC ln X C + n D ln X D )
and
ΔS mix = −8.3144
J
J
1
1
1
1
(1mol × ln + 1mol × ln + 1mol × ln + 1mol × ln ) = 46.1
mol × K
K
4
4
4
4
It is significant that the entropy of mixing is still positive and that the unites are J/K (since it is determined for a system, not
for a mole of any specific gas).
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