NAME:
PHGN341: Thermal Physics
Exam II - April 15, 2011 (Happy Tax Day!)
1. A throttling refrigerator uses HFC-134a as its coolant (see tables attached). It cycles as shown in the figure while
operating between, P1 = P4 = 1 bar and P2 = P3 = 10 bar. Recall that the compression step, 1 → 2, is isentropic and
the throttling step, 3 → 4, conserves enthalpy. The figure of merit for a refrigerator is the coefficient of performance,
COP, which is defined as the benefit (the amount of heat removed from the inside of the refrigerator) divided by the
cost (the work expended):
Qc
Qc
=
,
COP =
W
|Qh | − Qc
where energy conservation has been used.
P
(a) What is the temperature at point 2 in the cycle?
Liquid
Solution: The 1 → 2 step is adiabatic (isentropic); thus S1 = S2 .
From Table 2 we have S1 = 0.94 kJ/K. Extrapolate from
the data in Table 3 to find the temperature at point 2 that
has the same entropy as point 1:
S1 = αS40 + (1 − α)S50 .
Solving gives: α = (S1 − S50 )/(S40 − S50 ) = 0.0833.
This gives: T2 = α40o C + (1 − α)50o C = 49.17o C.
Vapor
P2 = P3
P1 = P4
2
3
4
Liquid + Vapor
1
V
(b) Find the coefficient of performance for this refrigerator.
Solution: As state above,
COP =
1
Qc
,
=
|Qh | − Qc
|Qh |/Qc − 1
where |Qh | = H2 − H3 and Qc = H1 − H4 . Points 1 and 3 are at the phase boundaries; so we can read their
values directly from Table 2: H1 = 231 kJ (pure gas) and H3 = 105 kJ (pure liquid). Because the throttling step
conserves enthalpy, we also have H4 = H3 = 105 kJ. From part (a) we have the interpolation constant needed to find
H2 = αH40 + (1 − α)H50 = 279.08 kJ. Plugging these into the expression for the coefficient of performance gives:
COP = 2.62.
2. When the volume of the liquid can be neglected compared to that of the gas phase, the Clausius-Clapeyron relation
for a liquid-gas phase transition becomes
dP
nL
nL
=
'
,
dT
T (Vg − Vl )
T Vg
where n is the number of moles of the substance and L is the molar latent heat of vaporization.
(a) Treating the gas as ideal, P Vg = nRT , where R = 8.315 J/K-mol, is the molar gas constant, find a differential
equation for the saturation pressure as a function of temperature. (Do not solve.)
Solution: Use the ideal gas law to write: Vg = nRT /P , then substitute this into the Clausius-Clapeyron relation to
find:
dP
L
= 2 P.
dT
T R
(b) The solution to your equation gives the liquid-vapor saturation pressure as a function of temperature which can be
written in the form
L
Ps (T ) = P0 e− RT ,
where T is the absolute temperature. At a pressure of Ps = 0.0317 bar one finds that water boils at T = 25 C. By
analyzing the energy needed for the phase transition at this temperature, one determines L ' 44.0 kJ/mol. From this
data find P0 .
1
L
Solution: From the above equation we have: P0 = Ps (T )e+ RT . Substituting the values above gives:
44×103
P0 = (0.0317bar)e+ 8.315·298 = 1.633 × 106 bar.
(c) The dew point is the temperature where the saturation vapor pressure is equal to the actual vapor pressure. Suppose
the temperature is 28 C and the relative humidity is 80%. What is the dew point (temperature)? (Recall that the
relative humidity is the ratio of the actual water vapor pressure to the saturation pressure. )
Solution: If the humidity is 80% at 28 C, then the absolute vapor pressure is: P = 0.80Ps (T = 28) = 0.03027 bar. The
temperature where this pressure saturates is:
T =−
L
= 297.2 K or 24.2 C.
R ln(P/P0 )
A
2
3. (a) Recall the particle-in-a-box result for the density of states in two-dimensions: d2 n = (2πh̄)
2 d p, where p is the
momentum and A is the area of the system. Use this to calculate the partition function for a single particle in two
p2
.
dimensions by integrating the boltzmann factor times the density of states over all p-space. For a free particle E = 2m
Solution: Perform the indicated integration:
Z
Z ∞
p2
p2
A
A
2
−β 2m
−β 2m
Z=
d
p
e
=
2π
dp
p
e
.
(2πh̄)2
(2πh̄)2 0
2
p
to obtain p dp =
Let x = β 2m
m
β dx.
Thus,
A m
Z = 2π
(2πh̄)2 β
Z
∞
dx e−x =
0
mA
.
2πβh̄2
(b) Using your result from part (a), find the the N-particle partition function assuming that all the particles are
indistinguishable.
Solution: For indistinguishable particles the N-particle partition function is: ZN = Z1N /N !.
(c) From N-particle partition function, find the Helmholtz free energy, F = −kT ln Z, as a function of the area, A,
temperature, T , number of particles, N , mass, m, and physical constants. Use the approximation, N ! ' N N e−N , to
simplify.
Solution: Using the stated approximation in the expression for the N-particle partition function gives:
mA N −N N
mA
F = −kT ln (
)
N
e
=
−N
kT
ln
−
ln
N
+
1
.
2πβh̄2
2πβh̄2
(d) From the Helmholtz free energy calculate the two-dimensional “pressure” (force per unit length), λ = −
Simplify to obtain the ideal gas law in two dimensions.
Solution: To simplify the derivative, break out the logarithm:
m
−
ln
N
+
1
.
F = −N kT ln A + ln
2πβh̄2
Thus,
λ=−
∂F
∂A
or, in standard ideal gas law form: λA = N kT .
2
= +N kT
N,T
1
,
A
∂F
∂A N,T .
Thermodynamic Tables
Substance
C (graphite)
C (diamond)
CaCO3 (calcite)
CaCO3 (aragonite)
Al2 SiO5 (kyanite)
Al2 SiO5 (andalusite)
Al2 SiO5 (sillimanite)
H2 O (l)
H2 O (g)
H2 (g)
O2 (g)
CH4 (g)
CO2 (g)
∆f H (kJ)
0.0
1.895
-1206.9
-1207.1
-2594.29
-2590.27
-2587.76
-285.83
-241.82
0.0
0.0
-74.81
-393.51
∆f G (kJ)
0.0
2.900
-1128.8
-1127.8
-2443.88
-2442.66
-2440.99
-237.13
-228.57
0.0
0.0
-50.72
-394.36
S (J/K)
5.74
2.38
92.9
88.7
83.81
93.22
96.11
69.91
188.83
130.68
205.14
186.26
213.74
CP (J/K)
8.53
6.11
81.88
81.25
121.71
122.72
124.52
75.29
33.58
28.82
29.38
35.31
37.11
V (cm3 )
5.30
3.42
36.93
34.15
44.09
51.53
49.90
Table 1. Thermodynamic properties of selected substances for one mole of the substance at standard temperature and
pressure.
T (◦ C)
-26.4
-18.8
-10.1
8.9
21.6
31.3
39.4
46.3
P (bar)
1.0
1.4
2.0
4.0
6.0
8.0
10.0
12.0
Hliquid (kJ)
16
26
37
62
79
93
105
116
Hgas (kJ)
231
236
241
252
259
264
268
271
Sliquid (kJ/K)
0.068
0.106
0.148
0.240
0.300
0.346
0.384
0.416
Sgas (kJ/K)
0.940
0.932
0.925
0.915
0.910
0.907
0.904
0.902
Table 2. Thermodynamic properties of saturated HFC-134a refrigerant for 1 kg of material at the boiling point temperatures for each pressure.
P (bar)
8.0
10.0
12.0
Property
H (kJ)
S(kJ/K)
H (kJ)
S(kJ/K)
H (kJ)
S(kJ/K)
Temperature
40
50
274
284
0.937 0.971
269
280
0.907 0.943
276
0.916
(◦ C)
60
295
1.003
291
0.977
287
0.953
Table 3. Thermodynamic properties of superheated HFC-134a refrigerant for 1 kg of material.
3