105 Physics Problem Set 11 Solutions
Problem 1 (3 pts)
(a) When bouncing o a wall perpendicular to the x-axis, only the x-component of the atoms velocity
changes its sign. Collisions with the other walls do not aect the x-component of the atoms velocity.
Hence the atom transfers to the wall orthogonal to the x-axis the amount of momentum equal to
p = 2mvx
during each collision; and it happens exactly once during the time interval equal to
t =
2L
.
vx
Consequently, the time averaged force is
F=
2
p
mvx
=
.
t
L
(b) Using the result from the previous part, the pressure of the gas on the wall orthogonal to the x-axis is
N
i=1
L2
p=
Fi
1
=3
L
N
2
mi vix .
i=1
Given that the direction of the atoms velocity is equally likely to point in any direction, we nd:
1
2
N
2
mi vix =
i=1
1
2
N
2
mi viy =
i=1
1
2
N
2
mi viz =
i=1
N
kB T .
2
This yields for the pressure
N
(kB NA )T .
NA
Recalling that the ideal gas constant is R = kB NA , we recover the ideal gas law.
pV =
Problem 2 (3 pts)
1. (Fermi 45.1) For the isothermic expansion at T1 = 400 K the amount of heat transferred to the gas
equals to the work done,
V2
= RT1 ln 5 = 5.3kJ .
V1
= const, for the compression at T2 = 300 K we have
Q1 = RT1 ln
Since the adiabatic process obeys T V k1
V2 /V1 = 5 as well. The amount of heat absorbed by the second source is
Q2 = RT2 ln
V1
= RT2 ln 5 = 4.0kJ .
V2
The net work done is
W = Q1 + Q2 = 1.3kJ
1
2. (Fermi 45.2) The maximum eciency is achieved when the engine works according to the Carnot
cycle. The upper temperature is T1 = 673K , and the lower temperature is T2 = 291K . Engines
eciency is
=
T1 T2
= 0.57 .
T1
3. (Fermi 45.3) Assuming that the heat is extracted in an isothermal process and using the reverse
Carnot cycle, the amount of work needed is
W =Q
T1 T2
= 0.22cal = 0.92 J .
T2
Here T1 = 311K is temperature of the environment, and T2 = 255K is the temperature of the body.
4. (Fermi 75.1) Change of the waters entropy during the heating process is
T2
S =
T1
mcV dT
T2
= 312cal/deg = 1.3kJ/K .
= mcV ln
T
T1
Dimensionless entropy change is therefore
S
= 9.4 1025 .
kB
Problem 3 (3 pts)
Note that the the rst compartment is at the constant volume, while the second compartment is at the
constant Let pressure.
(a) Q be the amount of heat transferred to the second compartment. Then for the rst compartment
we have
Q =
3
n1 R(T1 T ) .
2
For the second compartment:
Q =
3
5
n2 R(T T2 ) + p2 V2 = n2 R(T T2 ) .
2
2
Here we have used the ideal gas law for the second compartment,
p2 V2 = n2 RT2 .
(Recall that the pressure p2 is constant, so that p2 V2 = n2 RT2 .) Therefore, the nal temperature
T is found from the following equation:
5
3
n1 R(T T1 ) + n2 R(T T2 ) = 0 ;
2
2
or
T=
3n1 T1 + 5n2 T2
= 464 K .
3n1 + 5n2
2
(b) The amount of heat transferred is
Q =
3
15n1 n2
n1 R(T1 T ) =
(T1 T2 ) = 102 J .
2
2(3n1 + 5n2 )
(c) From the ideal gas law for the second compartment
p2 Ah = n2 RT2 .
The constant pressure in the second compartment is
p2 = patm +
mg
.
A
Here m is the mass of the piston, and A = 78.5cm2 is the cross section of the piston. These give
h =
n2 RT2
= 5.1cm .
patm A + mg
5
(d) Given that Q = 2 n2 RT and W = n2 RT , the fraction of heat converted to work is
W
2
=.
Q
5
Problem 4 (3 pts)
We nd that Vmin =
1000cm3
r 1
= 50cm2 and Vmax = 1050cm3 .
(a) At point 1 the intake air is at the external pressure and temperature. The number of moles of the gas
is
n=
p1 V1
= 0.0424 .
RT1
Equation for the adiabatic process for the gas is
p2 = p1 r = 71atm .
From the ideal gas law the temperature is
T2 = T1 r 1 = 1010 K .
For the constant-pressure process 2 3 the rst law of thermodynamics and the ideal gas law give
Q = pV +
1
nRT =
nRT .
1
1
Consequently, T3 = T2 + T = 1820 K . The volume is found as
V3 = V2
T3
= 90 cm3 .
T2
3
Transition 3 4 is an adiabatic process, and it is treated analogously to 1 2. Finally we arrive at
the following table:
p, atm
V , cm3
T, K
1
1.0
1050
298
2
71.0
50
1010
3
71.0
90
1820
4
2.28
1050
681
Note that T4 /T1 = V4 /V1 in accordance with the ideal gas law.
(b) The exhaust part of the cycle (4 1) is equivalent to taking away the amount of heat QC :
QC =
1
nR(T4 T1 ) = 337 J .
1
Given that QH = 1000 J , the net work done is
W = QH QC = 663 J .
(c) Thermal eciency of the engine is
=
W
= 0.663 .
QH
(d) 2400 rpm corresponds to f = 40 cycles per second. The power output of 8 cylinders is
P = 8f W = 15.9 kW = 213 hp .
4
5