Chapter 21
Questions: 1, 4, 8, 9, 11, 12, 13, 16, 19, 20, 22
Problems: 3, 7, 9, 11, 17, 21, 23, 25, 31, 39, 43, 45
Questions
1. The …rst law allows for energy to be added or subtracted to the system.
Conservation of energy is for an isolated system.
4. Heat is energy transferred to (or from) the system, while internal
energy is a measure of the systems internal energy typically characterized by
its temperature.
8. The system is not in quasi-static equilibrium and the gas doesn’t have
a well de…ned P and or V during this process.
9. The …rst law is basically the conservation of energy and thus applies
to the end states of an irreversible process. However an irreversible process
cannot be described continuously between these two end states.
11. No. Since it is quasi-static, the path (e.g. in a P-V diagram) is
continuous but not necessarily at constant temperature all along the path.
12. U is the same for both processes. However the gas in process A
does the more work than the gas in process B. Hence from the 1st law of
thermodynamics process A requires the larger heat transfer to the system.
13. Since the gas is in the same initial and …nal states, they have returned
to the same temperature. By the ideal gas law, they are at the same internal
energy. That being said two di¤erent processes could have required di¤erent
heat transfer with a corresponding di¤erent amount of work done by the gas.
16. Adiabatic expansion (or contraction) involves temperature changes,
speci…cally T V 1 = const.
19. A system that expands isothermally has heat added to maintain its
temperature. The work done by the gas is provided by the added heat.
A system that expands adiabatically has no heat added so that all the
work performed by the gas causes it to cool. The added heat in an isothermal
process allows that process to do more work.
A system that expands isobarically (constant pressure) must heat up to
satisfy the ideal gas law. Thus it has the most heat added allowing the gas
to do the most amount of work (of these three processes).
20. From the 1st law of thermodynamics Q W = U:At constant
volume there is no work done, W = 0. Hence at constant volume Q = U =
nCv T where V is constant and no work has been done.
22. In that limit Cv >> R and = Cp =Cv = (Cv + R) =Cv ' 1: Since
Cv >> R it requires more heat to change the temperature of this gas.
1
Problems
3. From the …rst law, Q = U +W: For this problem Q = 40 25 = 1000J:
During the expansion of the gas it does 750J of work. Hence U = 250J:
7. From the …rst law, Q = U + W: Thus
U
W
+
;
t
t
U
=
+ 400 106 ;
t
m
= 90 106 W =
c T:
t
Q
=
t
50
106
9:8
U
t
Thus the change in temperature of the water is
T =
90 106 J=s
90:
=
= :0215 C.
6
10 kg=s (4184J=kgC)
4184
9. States A, B, and C are de…ned by A : P = P1 ; V = V1 ; B : P =
2P1 ; V = 2V1 ; C : P = 2P1 ; V = V1 :
For a gas going from A to C we have V = 0; hence
Z V1
WAC =
P dV = 0:
V1
For a gas going from C to B , P = 2P1 ; hence
Z 2V1
Z 2V1
WCB =
P dV = 2P1
dV = 2P1 V1 :
V1
V1
Hence the total work is W = WAC + WCB = 2P1 V1 :
11. For an isothermal expansion the pressure of a gas is given by P V =
Pi Vi = :nRT = :3 8:314 300J = 748: 3J. Thus the work done in expanding
to 5 times its original volume is
Z 5Vi
nRT
dV = 748: 3 ln 5 = 1:204kJ.
W =
V
Vi
17. The work required to isothermally compress a gas (the negative of
the work done by the gas) to half its original volume is
W = nRT ln 2 = 600 ! nRT = 600= ln 2:
2
The work required to isothermally compress the same gas to 1/10 of its
original volume is
W = nRT ln 10 = 600
ln 10
= 1:993kJ
ln 2
21. (a) Along an adiabat ( = 1:4) the pressure at B is found from
51:4 = 571kP a
PB = PA (VA =VB ) = 60kP a
(b) The net work done on the gas is the negative of the work done by the
gas through the entire cycle. The work done on the gas during the adiabatic
compression is
WBA =
WBA =
PB VB
PA VA
571kP a
=
1
1L 60kP a
:4
5L
271
= 677:5J:
:4
The work done on the gas during the constant pressure part of the cycle is
WCA =
PA V =
60kP a
4L =
240J:
The net work done on the gas is the sum of these two quantities or
Wnet = (677:5
240) J = 437:5J:
23. A gas undergoing adiabatic expansion satis…es
Ti Vi
1
= Tf Vf
1
,
Tf = Ti (Vi =Vf )
1
.
For a compression ratio of Vi =Vf = 8:5 and = 1:4 the …nal temperature is
given by
Tf = 303K (8:5):4 = 713: 20K = 440: 20 C:
25. For Tf =Ti = 2 in an adiabatic compression ( = 1:4), the volumes
must satisfy
(Vi =Vf ) 1 = (Vi =Vf ):4 = 2
Vi =Vf = 22:5 = 5:66
Vf = :177Vi
3
31. (a) An ideal gas with = 5=3 has an initial pressure in state A
of PA = 250kP a and an initial volume of VA = 1m3 . The gas is adiabatically expanded to a volume in state B of VB = 3m3 : The pressure at B is
determined via
PB VB = PA VA ;
PB = PA (VA =VB ) = 250 (1=3)5=3 kP a = 40kP a.
(b) If the gas were allowed to expand isothermally to state C with a
volume of VC = VB = 3m3 ; then it pressure at C would be given by
PC VC = PA VA ;
PC = PA VA =VC = PA =3 = 83:3kP a.
The net work done on the gas is given by the area inside the P-V curve in
…gure 21-29. This expressed as
Z B
Z B
VA
VA
dV;
(Piso Pad ) dV = PA
W =
V
A
A V
V
1
1
W = PA VA ln 3 + A
;
1
1 (3VA )
VA 1
1
1
W = PA VA ln 3 +
1
= PA VA (1:0986 :7789) ;
1 3 1
W = :3197 (250) kP am3 = 79:9J
39. From the …rst law of thermodynamics
Q=
U + W:
At constant pressure the work done is easily found from the ideal gas law,
Z
W = P dV = P V = nR T:
Air is made up primarily of diatomic molecules, hence its molar speci…c heat
is CV = 5R=2: Thus the heat required to heat 5L of air under 100kP a of
pressure from 0 C to 20 C is
7
Tf Ti
7
Q = nRTi
= 100kP a5
2
Ti
2
10 3 m3
4
20
7
100
= 100J
= 128:2J:
273
2
273
43. For a mixture of gasses we have
5
3
Q = n m R T + n d R T = nCv T;
2
2
where m and d are the relative fractions of monatomic and diatomic gasses
respectively.
Now
R
Cp = Cv = Cv + R ! Cv =
:
1
For a mixture of gasses where = 1:52
+
3
5
m+
2
2
m
Solving for
m
= 1
1
=
= 1:923
:52
d
d
we …nd
m
= det
1
1
1:923 2:5
= :577 = 57:7%
45. Under adiabatic expansion of a mixture of argon (Cv = 3=2) and
oxygen (Cv = 5=2) experiences a pressure drop of 1=3 while doubling its
volume. Under adiabatic expansion
Pi Vi
= Pf Vf ; ! Pi =Pf = (Vf =Vi ) ;
ln 3
ln Pi =Pf
=
= 1:58496:
=
ln Vf =Vi
ln 2
Since CP = Cv + R we have Cv = Cv + R or Cv = R= (
1) : This means
the speci…c heat for this mixture is Cv = 1:71R: The internal energy for the
mixture is
5
3
= nCv T = n1:71RT;
U =n
o RT + a RT
2
2
where n is the total number of moles for the gas mixture and
the relative fractions of oxygen and argon respectively. Thus
+
5 o+3
o
Solving for
a
= 1; and
= 3:42:
a
a
we …nd
a
= :79 = 79%:
5
o
and
a
are