Download Solution

Physics H7B — Siddiqi (Fall 2011)
Midterm #1 Solutions
Charles Melby-Thompson
Problem 1: One mole of an ideal diatomic gas is caused to pass through the
cycle shown on the P-V diagram. 1 → 2 is an isothermal process; 2 → 3 is a
constant volume process; and 3 → 1 is an adiabatic process. Assume that the
thermal properties of the gas are correctly described by classical electrodynamics.
(a) What is the ratio Cp /Cv for the gas?
Solution: For one mole of ideal gas, Cp = Cv + R. Since the gas’s thermal properties are described correctly by classical thermodynamics, the equipartition theorem
applies to the vibrations of the gas as well as its rotations. Because the equipartition theorem applies to the potential as well as kinetic energy of the vibrations,
9
the number of degrees of freedom is 7. Thus γ = 7+2
7 = 7.
(b) Calculate P2 and P3 .
Solution: Process 1 → 2 is isothermal. Since the temperature is proportional to
P V , we find
P1
P1 V1 = P2 V2 = P2 (3V1 )
=⇒
P2 =
.
3
3 → 1 is adiabatic, so
P1 V1γ = P3 V3γ = P3 (3V1 )γ
=⇒
P3 =
P1
.
3γ
(c) Find the temperatures at 1, 2, and 3.
Solution: From the ideal gas law T1 =
1 V1
Finally, T3 = P3RV3 = 3Pγ−1
R.
P1 V1
R .
Since 1 → 2 is isothermal, T2 = T1 .
(d) Calculate |Qin | and |Qout |. During which legs of the cycle are these non-zero?
Solution: During 1 → 2 the temperature is constant, and thus the energy and P V
are, too. The gas expands and does work, so heat must flow in to keep U constant.
The first law then says
Z 2
Z V2
V2
P1 V 1
dV = P1 V1 log
= P1 V1 log 3.
Qin = Wby =
P dV =
V
V
1
1
V1
In 2 → 3 the gas does no work and its energy goes down, thus it loses heat during
this process. The energy for the gas is Cv T = 72 RT , so the first law gives
7
7
1
7
Qout = −∆U = R(T2 − T3 ) = P1 V1 1 − γ−1 = P1 V1 (1 − 3−2/7 ).
2
2
3
2
There is no heat transferred during 3 → 1 since it is adiabatic.
(e) Find the work done W and the efficiency η.
1
2
Solution: After one cycle ∆U = 0 and therefore the first law tells us that
7
−2/7
W = Qin − Qout = P1 V1 log 3 − (1 − 3
) .
2
The efficiency is then
η=
W
7(1 − 3−2/7 )
=1−
≈ 14%.
Qin
2 log 3
Problem 2: Consider a Carnot engine where the working fluid is 0.3 kg of water.
The engine operates between a high temperature T1 = 473K and a low temperature
T2 = 373K. During a cycle, the water undergoes a phase transition between liquid
and vapor. The cycle is shown in the P-V diagram below. Legs 1-2 and 3-4 are
isotherms at T1 and T2 , respectively, and legs 2-3 and 4-1 are adiabatic expansion
and contraction, respectively. The dashed line represents the boundary of the
liquid-vapor equilibrium regions. What is the work done by the engine in one
cycle?
Solution: The engine is a Carnot engine, so Qout = TT21 Qin . Therefore
T1
W = Qin − Qout =
− 1 Qout .
T2
Heat is expelled only in 3 → 4. From the graph we see that this process consists
entirely of an isothermal phase change from vapor to liquid at the boiling point T2 .
Therefore the heat expelled equals mLv .
It follows that
W =
100K
T1
− 1 mLv =
(0.3kg)(2260 kg/kJ) ≈ 182 kJ.
T2
373K
Problem 3: Consider an ideal gas undergoing a polytropic process described by
the characteristic relation P V δ = constant.
(a) The diagram above describes the expansion of an ideal gas from State 1 to
different final states labeled ’State 2’. Compare (using <, >, =) the values of P , V ,
and T in State 1 and State 2 for each process drawn. In the regions defined by the
roman numerals i, ii, iii, where is Q > 0 and where is Q < 0?
Solution:
process
isobaric
isothermal
adiabatic
isochoric
P
1=2
1>2
1>2
1>2
V
1<2
1<2
1<2
1=2
T
1<2
1=2
1>2
1>2
Q
i >0
ii > 0
iii < 0
(b) What is the value of δ for the four cases plotted in the P-V diagram? We can
compare the value of δ to the specific heat ratio γ = Cp /Cv . In which regions is
δ > γ, and where is δ < γ?
3
Solution:
Isobaric: P = P V 0 = c =⇒ δ = 0
Isothermal: nRT = P V = P V 1 = c =⇒ δ = 1
Adiabatic: P V γ = c =⇒ δ = γ
Isochoric: P V δ = c is the same as P 1/δ V = c1/δ = c0 . Therefore V = P 0 V =
P 1/δ V = c0 =⇒ 1/δ = 0, i.e. δ = ∞.
The adiabatic process is the boundary between δ < γ and δ > γ, and so δ < γ in i
and ii, and δ > γ in iii.
(c) For a general polytropic process, derive the temperature of the ideal gas in
terms of thermodynamic variables. If the gas is expanding, for what values of δ
does the temperature increase? In which regions does the temperature decrease?
Solution: If P V δ = c is constant then T V δ−1 = P V δ /nR is also constant. Thus
T = c/nRV δ−1 ∝ 1/V δ−1 .
Therefore, as V increases: T increases if δ < 1; and T decreases if δ > 1.
(d) In class, we derived the heat capacity at constant volume and constant pressure
for an ideal gas in terms of γ. We can generalize the heat capacity for a general
process using the characteristic relation dQ = N C dT . Find C for a general polytropic process. Show that it reduces to the values derived in class for isobaric and
isochoric processes.
Solution: For an ideal gas γ = 1 + kCBv (Cv is the molecular heat capacity) and so
kB
Cv = γ−1
. An ideal gas also has U = N Cv T .
In a polytropic process with T V δ−1 = c we have
V
d(T V δ−1 )
1 V
= (δ − 1)dV + dT = 0 =⇒ dV = −
dT.
T V δ−2
T
δ−1T
Therefore
dQ = dU + P dV = N Cv dT −
and
1 PV
dT = N kB
δ−1 T
1
1
−
γ−1 δ−1
dT
1
δ−γ
1
C = kB
−
= kB
.
γ−1 δ−1
(δ − 1)(γ − 1)
For Cv (δ = ∞) and Cp (δ = 0) this gives the standard values
Cv =
kB
γ−1
Cp =
−kB γ
kB γ
=
= Cv + kB .
1−γ
γ−1
Problem 4:
In order to compensate for deviations caused by temperature
changes, a pendulum clock built in the nineteenth century for an astronomical
observatory uses a large cylindrical glass tube filled with mercury as a pendulum
bob. The tube is held by a brass rod and bracket whose combined length, measured
from the point of suspension of the pendulum, is `. Neglecting the mass of the brass
and glass, and neglecting the expansion of the glass, find an algebraic expression
(materials specific constants OK) for the height h the mercury must be filled to so
4
that the center of mass of the mercury is to remain at a fixed distance from the
point of suspension, regardless of temperature.
Solution: The center of mass is h/2 above the bottom of the pendulum bob and
so the distance from the pivot is r = ` − h/2. Since the volume of the mercury is
Vm = πr2 h, if the temperature now changes by ∆T then
Vm
βm V ∆T
∆h = ∆
=
= βm h∆T
2
πr
πr2
as the glass does not expand; and ∆` = αb `∆T . Thence
h
1
∆T,
∆r = ∆` − ∆h = αb ` − βm
2
2
and for this to vanish we must have
2αb `
h=
βm
with αb the coefficient of linear expansion for brass and βm that of volume expansion
for mercury.