Cornell University, Physics Department
PHYS-3341 Statistical Physics
Fall 2014
Prof. Itai Cohen
Solutions to Problem Set 5
David C. Tsang, Woosong Choi
5.1
Refrigeration
Reif §5.22: Refrigeration cycles have been developed for heating buildings. The
procedure is to design a device which absorbs heat from the surrounding earth
or air outside the house and then delivers heat at a higher temperature to the
interior of the building. (Such a device is called a “heat pump.”)
(a) If a device is used in this way, operating between the outside absolute temperature To and an interior absolute temperature Ti , what would be the
maximum number of kilowatt-hours of heat that could be supplied to the
building for every kilowatt-hour of electrical energy needed to operate the
device?
(b) Obtain a numerical answer for the case that the outside temperature is 0◦ C
and the interior temperature is 25◦ C.
(a) The change in total entropy is
∆S =
We want to know what
Qi
W
Qi Qi − W
−
≥0
Ti
To
is, so we move terms around and get
Qi 1
1
1
−
≥−
W Ti To
To
Thus for To < Ti ,
Qi
Ti
≤
W
Ti − To
(b) Substituting the values Ti = 298K and Ti = 273K we get
Qi
≤ 11.9
W
2
1
5.2
Gas Engine
Reif §5.26: A gasoline engine can be approximately represented by the idealized cyclic process abcd shown in the accompanying diagram of pressure p versus
volume V of the gas in the cylinder. Here a → b represents the adiabatic compression of the air-gasoline mixture, b → c the rise in pressure at constant volume
due to the explosion of the mixture, c → d the adiabatic expansion of the mixture
during which the engine performs useful work, and d → a the final cooling down
of the gas at constant volume.
Assume this cycle to be carried out quasi-statically for a fixed amount of ideal
gas having a constant specific heat. Denote the specific heat ratio by γ ≡ cp /cv .
Calculate the efficiency η (ratio of work performed to heat intake Q1 ) for this
process, expressing your answer in terms of V1 , V2 , and γ.
First, for the heating and cooling processes we can write
Q1 = νcv (Tc − Tb )
Q2 = νcv (Td − Ta )
Since over a cycle abcd the gas returns to its original state, Q1 = Q2 +W . Thus, the efficiency
is
Q1 − Q2
Q2
Td − Ta
W
=
=1−
=1−
η=
Q1
Q1
Q1
Tc − Tb
For adiabatic proccess, pV γ is constant and for ideal gas pV = νRT , T V γ−1 is also constant.
Thus we can relate Ta and Td to Tb and Tc respectively.
γ−1
V1
Ta = Tb
V2
γ−1
V1
Td = Tc
V2
Substituting this in the equation of η we get
Tc − Tb
η =1−
Tc − Tb
V1
V2
γ−1
=1−
V1
V2
γ−1
2
2
5.3
Triple Point I
Reif §8.2: The vapor pressure p (in millimeters of mercury) of solid ammonia
is given by ln p = 23.03 − 3754/T and that of liquid ammonia by ln p = 19.49 −
3063/T .
(a) What is the temperature of the triple point?
(b) What are the latent heats of sublimation and vaporization at the triple
point?
(c) What is the latent heat of melting at the triple point?
(a) Triple point is the point where the phase transition between all three phases occur. For
that to happen, the phase transition curve between solid-vapor and liquid-vapor must
coincide at a point. Equating the equations we get
3063
3754
= 19.49 −
TC
TC
3754 − 3063
23.03 − 19.49 =
TC
691
' 195.2
TC =
3.54
ln p = 23.03 −
(1)
Thus,
TC = 195.2K
(b) From the Clausius-Clapeyron equation (Reif §8.5.10)
L12
dp
=
dT
T ∆V
Assuming that the ammonia vapor obeys the ideal gas law and that the volume of
the solid and the liquid is negligibly small compared to the volume of the gas, we can
deduce
l
ln p = −
+ constant
RT
Thus, from the given equations we can find that
lsublimination = 3754R ' 3754K × 8.31J/mol · K ' 3.12 × 104 J/mol
lvaporization = 3063R ' 3063K × 8.31J/mol · K ' 2.55 × 104 J/mol
(c) At the triple point,
lsolid→liquid = T (Ssolid − Sliquid )
= T (Ssolid − Sgas + Sgas − Sliquid )
= lsublimination − lvaporization
3
Thus, the latent heat of melting is
lmelting ' (3.12 − 2.55) × 104 J/mol = 5.7 × 103 J/mol
Box
5.4
Latent Heat
Reif §8.7: The molar latent heat of transformation in going from phase 1 to
phase 2 at the temperature T and the pressure p is l. What is the latent heat of
the phase transformation at a slightly different temperature (and corresponding
presusre), i.e., what is (dl/dT )? Express your answer in terms of l and the molar
specific heat cp , coefficient of expansion α, and molar volume v of each phase at
the original temperature T and pressure p.
dS2 dS1
dl
=
−
T + (S1 + S2 )
l = (S2 − S1 )T
⇒
dT
dT
dT
We also have
Cp
∂V
Cp
dS
dp Cp
−V αl Cp
dS =
dT −
dP = −V αdp + dT
⇒
= −V α
+
=
+
T
∂T
T
dT
dT
T
T ∆V
T
−V2 α2 l Cp2 V1 α1 l Cp1
1
V1 α1 − V2 α2
dl
l
=
+
+
−
+
T + = Cp2 − Cp1 + l
dT
T ∆V
T
T ∆V
T
T
T
V2 − V1
2
4
5.5
Tie Lines
Bowley and Sanchez §11.6: Suppose that the thermodynamic potential for a system with
a conserved quantity φ is of the form
Φ(φ) = Φ(0) + c(φ2 − φ2a )2
What it the slope of the double tangent to the curve? Obtain φ∗1 and φ∗2 , the
values of φ where the double tangent touches the curve. What are the spinodal
limits, φ¯1 and φ¯2 , for this thermodynamic potential? Show that the spinodal
limits lie between φ∗1 and φ∗2 . For what value of φ is the system in a metastable
state?
(a) First, we will find the double tangent values φ∗1 and φ∗2 . For the two values of the double
tangent touching the curve, it should satisfy
∂Φ
∂Φ
Φ(φ∗2 ) − Φ(φ∗1 )
=
=
∂φ φ∗
∂φ φ∗
φ∗2 − φ∗1
1
2
From the first part of the equality, we get
4cφ∗1 (φ∗1 2 − φ2a ) = 4cφ∗2 (φ∗2 2 − φ2a )
(φ∗1 − φ∗2 )(φ∗1 2 + φ∗1 φ∗2 + φ∗2 2 − φ2a ) = 0
From this, φ∗1 = φ∗2 or φ∗1 2 + φ∗1 φ∗2 + φ∗2 2 − φ2a = 0. But, we don’t want φ∗1 = φ∗2 since it
would not lead us to a double tangent. Now, take the average of the first two derivatives
in the equation it should still be the same as the right hand side of the equation, since
all three terms are equal. That is,
"
#
∂Φ
∂Φ
Φ(φ∗2 ) − Φ(φ∗1 )
1
+
=
2
∂phi φ∗
∂phi φ∗
φ∗2 − φ∗1
1
2
φ∗1 2 (φ∗1 2 − 2φa 2 ) − φ∗2 2 (φ∗2 2 − 2φa 2 )
+
−
= c
−
φ∗1 − φ∗2
1
2(φ∗1 + φ∗2 )(φ∗1 2 − φ∗1 φ∗2 + φ∗2 2 − φ2a ) =
(φ∗ 2 − φ∗2 2 )(φ∗1 2 + φ∗2 2 − 2φ2a )
∗
φ1 − φ∗2 1
2cφ∗1 (φ∗1 2
φ2a )
2cφ∗2 (φ∗2 2
φ2a )
Now, we since φ∗1 6= φ∗2 we can divide φ∗1 2 − φ∗2 2 on the right hand side by φ∗1 − φ∗2 and
simplify
(φ∗1 + φ∗2 )(φ∗1 2 − 2φ∗1 φ∗2 + φ∗2 2 ) = (φ∗1 + φ∗2 )(φ∗1 − φ∗2 )2 = 0
Thus, φ∗1 = −φ∗2 . By φ∗1 2 + φ∗1 φ∗2 + φ∗2 2 − φ2a = 0 this becomes φ∗1 2 = φ2a .
φ∗1,2 = ±φa
5
(b) Now we want to find the spinodal limits for this potential. Spinodal limits are the values
2
where ∂∂φΦ2 = 0.
∂ 2Φ
= 4c(φ2 − φ2a + 2φ2 ) = 4c(3φ2 − φ2a ) = 0
∂φ2
Solutions to this equation is
φa
φ̄1,2 = ± √
3
Since φ∗ = ±φa from (a), we see that the spinodal limit values φ̄1,2 lie between φ∗1 and
φ∗2 .
(c) Metastable states are the states with the second derivative of the potential negative
while the phase parameter remaining within the double tangent contact points. Since
2
the second derivative is ∂∂φΦ2 = 4c(3φ2 − φ2a ), for values 3φ2 < φ2a we get negative second
√
√
derivative. For φ values from −φa to −φa / 3 and from φa / 3 to φa , we get positive
second derivative which means that the states are metastable.
2
6