Download Chem 3411 test #2 solutions – out of 30 marks. Problem 1 – 10

Chem 3411 test #2 solutions – out of 30 marks.
Problem 1 – 10 marks
The inside of an icemaking machine is maintained at 270 K (26.6 F), and heat is pumped
into a hot kitchen at 303 K (86 F). Water at 288 K (59 F) is made into ice at 273 K (32 F). In
actual (irreversible) operation, the electrical work required is 1.35 times that for reversible
operation. What electrical power (in watts) must be supplied to produce 0.500 kg of ice per
minute?
Data:
molar mass of H = 1 g/mol
molar mass of O = 16 g/mol
constant pressure heat capacity of liquid water = 75.3 J/(K mol)
enthalpy of fusion for H2 O at 273 K = 6.008 kJ/mol
one watt = one joule per second
solution
The icemaking machine is a heat pump operating in reverse. TC =270 K, TH =303 K.
0.500 kg of ice / (18 g/mol) = 27.75 moles of ice. How much heat must be extracted to
go from H2 O(liquid, 288 K) → H2 O(solid, 273 K) at constant pressure? Cooling the liquid
takes q=CP ∆T = -1129 J/mol. Then turning the liquid to ice takes an additional -6.008
kJ/mol. So for the complete process we need to extract 7.7137 kJ/mol, which becomes
198 kJ given how many moles we have. Hence we must pump 198 kJ of heat from TC to
TH . Thus qC =198 kJ. Now, (-w/qH )=(TH -TC )/TH = 0.1089 if reversible. But qH =-w-qC .
Solving for w gives w=24.2 kJ. But in actual operation w=(1.35)(24.2 kJ) = 32.7 kJ. To
produce this amount of ice (ie. transfer this amount of heat) every minute, the electrical
power supplied must be 32.7 kJ/60s = 544 W.
Problem 2 – 10 marks
A 0.250 mole sample of liquid water at 283 K (the system), when placed in a freezer
maintained at 268 K and 1 bar (the surroundings), spontaneously freezes. The temperature
of the water is measured as heat is removed from it at a constant rate, with the results
plotted below.
(a) use the plotted data to calculate ∆H0fus and ∆S0fus (both per mole) for the water at
its normal freezing point.
solution
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∆H0fus =(1690 - 188) J / 0.25 mol = 6.008 kJ/mol
∆S0fus =∆H0fus /Tfus = (6.008 kJ/mol)/(273 K) = 22 J/(K mol)
(b) use the plotted data to calculate CP (per mole) for liquid water
solution
use the liquid branch: CP =188 J / 10 K /0.25 mol = 75.2 J/(K mol)
(c) Calculate ∆Ssystem (for the actual 0.250 mole sample) for the process: H2 O(liquid,
283 K) → H2 O(solid, 273 K).
solution
Z
273
∆S =
283
CP (T )
dT − ∆Sfus
T
(1)
=75.2 J/(K mol) ln(273/283)-22 J/(K mol) times 0.25 mol = -6.2 J/K
Note that fusion is for solid to liquid, and we are going from liquid to solid.
(d) Explain why a portion of the plotted curve is horizontal.
The heat is causing a phase transition rather than a temperature change.
Problem 3 – 5 marks
For one mole of H2 O(liquid) → H2 O(gas), occuring at constant temperature and pressure,
we know that ∆Gvap =0 at P=1 atm and T=373.15 K (the boiling point of water at 1 atm).
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This result occurs because G=H-TS and ∆Svap =∆Hvap /Tvap .
We saw in class that, always at P=1 atm, at T=372.15 K, ∆G=108 J/mol, and that
at T=374.15 K, ∆G=-108 J/mol. What do these two ∆G values mean in terms of the
spontaneity of the process?
solution
∆G=108 J/mol indicates this is not spontaneous, which makes sense since below the
boiling point, we shouldn’t get conversion of liquid to vapor.
∆G=-108 J/mol indicates that this is spontaneous, which makes sense since we are above
the boiling point.
What shifts to cause G to be so sensitive to temperature? Is it primarily an enthalpy or
an entropy effect? Give as mathematically rigorous an answer as possible.
solution
∂G
∂T
= −S
(2)
P
Hence its an entropy effect.
Problem 4 – 5 marks
The first law of thermodynamics says that energy is conserved. The second law, however,
implies that heat and work are not equivalent forms of energy. Explain. In answering,
include a consideration of whether heat can be converted to work, and vice versa.
solution
work can always be converted to heat, but heat cannot be converted to work with 100%
efficiency.
Problem 5 – 5 marks
There are four Maxwell relations, one each for U, H, A, and G. Derive one of these (your
choice) and explain how you (finally) arrive at the Maxwell relation. In other words, what
result from calculus allows you to write down the Maxwell relation?
solution
The result from calculus is the equality of mixed partials, irrespective of the order.
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