NAME:
KEY
Section Number: 10
CHEMISTRY 418, Fall, 2010(10F)
Midterm Examination #1, October 7, 2010
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) An experimental setup has two compartments (V1 = 1 L, V2 = 3 L), connected by a valve. DO NOT WRITE
Initially, the valve is closed and 50 g of Ar gas is placed at room temperature into compartment one, IN THIS SPACE
100 g of Xe gas is placed into compartment two. Use ideal gas approximation to determine the total
pressure in the setup once the valve is open and the system attains equilibrium at room temperature.
p. 1________/20
p. 2________/20
Answer:
For each gas: PV = nRT
p. 3________/20
p. 4________/20
50 g
50 g
100 g
100 g
n1 ( Ar ) =
=
= 1.252 mol ; n 2 ( Xe) =
=
= 0.762 mol p. 5________/20
M ( Ar ) 39.948 g / mol
M ( Xe) 131.293 g / mol
=============
V = V1 + V2 = 4 L
At room temperature:
P1 ( Ar ) =
p. 6_______/5
(Extra credit)
=============
TOTAL PTS
n1 ( Ar ) RT 1.252 mol × 8.3144 J / mol / K × 298.15 K
J
=
= 775907.7 3 = 775907.7 Pa
3
V
m
m
4L × 0.001
L
P2 ( Xe) =
/100
n 2 ( Xe) RT 0.762 mol × 8.3144 J / mol / K × 298.15 K
J
=
= 472237.8 3 = 472237.8 Pa
3
V
m
m
4L × 0.001
L
Ptotal = P1 ( Ar ) + P2 ( Xe) = 775907.7 Pa + 472237.8 Pa = 1248145.5 Pa = 12.318 atm
Although this is a pretty high pressure, the gases used are very close to being perfect so ideal gas law is readily applicable.
Also, since the ideal gas approximation is used, there is no need to account for possible interactions between the molecules
of different kind.
NAME:
CHEM 443, Midterm Exam #1, Fall, 2005, page 2
2. (20 points)
a)
(4 points). State the Zeroth Law of thermodynamics
Answer: Two systems that are separately in thermal equilibrium with a third system are also in thermal equilibrium with one
another
b)
(4 points). Explain in a sentence or two the improvements of the van der Waals equation of state compared to the
ideal gas equation of state
Answer: van der Waals equation of state introduces the size of the molecules, which accounts for repulsive interactions, and
describes the attractive forces (constant a). This combination allows for a better prediction of gas behavior at high pressures
and low temperatures and theoretically predicts the existence of a liquid phase (albeit more qualitatively than quantitatively
describes it).
c)
(4 points). Explain in one or two sentences the difference between reversible and irreversible processes
Answer: If the direction of a process can be changed by an infinitesimal force, this process is reversible (also a process that
can be described as moving between equilibrium states). If an infinitesimal force does not change the direction of a process,
such process is called irreversible (also spontaneous).
d)
(4 points). State the Third Law of thermodynamics
Answer: The entropy of a pure, perfectly crystalline substance is zero at 0 K.
e)
(4 points). Explain why reaction enthalpies depend on temperature. You can use formulas or graphics if it would help
your explanation.
Answer: According to the equation describing temperature-dependence of enthalpy of a reaction on page 5-10 of the Blue
Book, the value of enthalpy depends on the heat capacities of reactants and products. These values can obviously have
different temperature dependences, as illustrated by Shomate equation, which means that the enthalpy change for a
reaction generally depends on temperature.
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NAME:
CHEM 443, Midterm Exam #1, Fall, 2005, page 3
3. (20 points) Prove that the process of adiabatic irreversible expansion described by the Joule-Thompson experiment is
isenthalpic (ΔH = 0). Show all your work clearly. (Hint: you can start with the First Law of thermodynamics and the
definition of enthalpy to describe the process of adiabatic irreversible expansion)
1st Law : ΔU = q + w , but since the process is adiabatic, q = 0;
Definition of enthalpy: H = U + PV
Thus,
0
V2
V1
0
ΔU = w = wleft + wright = − ∫ P1 dV − ∫ P2 dV = − P2V2 + P1V1
ΔH = ΔU + Δ ( PV ) = − P2V2 + P1V1 + ( P2V2 − P1V1 ) = 0
The enthalpy for the process does not change or the process is isenthalpic.
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NAME:
CHEM 443, Midterm Exam #1, Fall, 2005, page 4
4. (20 points). Calculate absolute molar entropy of diamond at 500°C and 1 bar. Use any information available in the
Blue Book and assume that the heat capacity of diamond can be approximated as a constant within the temperature
interval studied.
At constant pressure:
S
Θ
m , 773.15 K
=S
Θ
m , 298.15 K
773.15 K
+
∫
298.15 K
C p ,m (C , diamond )
T
dT = 2.4
773.15 K
J
J
J
ln
+ 6.1
= 8.2
mol × K
mol × K 298.15 K
mol × K
Although the number is very small, you can see that it more than triples at this high temperature. This means that the solid
diamond is very well ordered at room temperature but its disorder should affect its properties significantly at 500°C.
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NAME:
CHEM 443, Midterm Exam #1, Fall, 2005, page 5
5. (20 points). Use Hess’ Law and any enthalpies of formation available in Table 5.8 to calculate the standard enthalpy
Θ
( ΔH rzn ) for a reaction of partial oxidation of sucrose:
C12 H 22 O11 ( s ) + 6O2 ( g ) → 2C 2 H 5 OH (l ) + 8CO2 (aq) + 5H 2 O (l )
where CO2 (aq ) is CO2 dissolved in water (see below)
The following reactions can be used:
1)
C12 H 22 O11 ( s) + 12O2 ( g ) → 12CO2 ( g ) + 11H 2 O(l ), ΔH I = −5639.7 kJ (1) This is a combustion reaction
2)
CO2 ( g ) → CO2 (aq ), ΔH II = −19.4 kJ (2) Dissolution of CO2 in water.
Answer:
Θ
ΔH rzn
= ΔH I + 8 × ΔH II − 4ΔH Θf (CO2 ( g )) + 2ΔH Θf (C 2 O5 OH (l )) − 6ΔH Θf ( H 2 O(l ))
Θ
ΔH rzn
= −5639.7 kJ / mol + 8 × (−19.4 kJ / mol ) − 4 × (−393.5
kJ
kJ
kJ
) + 2 × (−277.6
) − 6 × (−285.8
)
mol
mol
mol
The stoichiometric coefficients have the units of mol but as long as the reaction used is balanced, the units of the enthalpy
of the reaction will be acceptable as either kJ or as kJ/mol for a full credit.
Θ
ΔH rzn
= −5639.7 kJ / mol − 155.2 kJ / mol + 1574
kJ
kJ
kJ
kJ
− 555.2
+ 1714.8
= −3061.3
mol
mol
mol
mol
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NAME:
CHEM 443, Midterm Exam #1, Fall, 2005, page 6
6. (5 points, extra credit) Calculate the maximum work that can be done by expanding 5 moles of Argon gas
(adequately described as an ideal gas) from 10 to 40 liters at room temperature. Show all your work clearly.
Maximum work can be achieved by a reversible process. Room temperature process means that it is isothermal.
For ideal gas in an isothermal process:
V2
V2
V
nRT
J
40
dV = − nRT ln 2 = −5 mol × 8.3144
× 298.15K × ln
= −17182.69 J = −17.18269 kJ
V
V1
mol × K
10
V1
w = − ∫ PdV = − ∫
V1
The significance of the negative sign is that the work is performed by the system on the surroundings.
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