# Chapter 12 – The Laws of Thermodynamics ```Chapter 12 – The Laws of Thermodynamics
This chapter deals with the laws of thermodynamics.
Work in thermodynamics processes
V
A
A gas in a cylinder can be compressed by pushing
on a piston, as shown to the right. The work done
on the gas is given by
F
V
W  Fx  PAx  PV
x
If the gas is compressed, then V is negative and
the work done is negative. If the gas expands (as free gas expansion), then the work done
by the gas is positive.
Isobaric process
In an isobaric process the pressure is constant. Thus, the work done is
W  PV  P(V f  Vi )
Isovolumeric process
In an isovolumeric (isochoric) process the volume is constant. Thus,
W  PV  0
If the pressure changes during the compression or expansion, then the work done on the
substance is the area under the P versus V curve.
Isothermal process
It can be shown (using calculus) that the work done in an isothermal compression is
W  nRT ln( V f / Vi )
isobaric
P
f
isovolumetric
i
P
isothermal
i
P
i
f
V
f
V
V
1
First Law of Thermodynamics
The first law of thermodynamics relates the change in internal energy to the heat
absorbed and the work done on a substance. It is essentially a statement of conservation
of energy.
U  U f  U i  Q NET  WNET
Q is positive if heat is absorbed and is negative if heat is lost by the system.
For a monatomic ideal gas, U  32 Nk BT  32 nRT . Thus, for an isothermal expansion or
compression, U  32 nRT  0 , and Q = W.
An adiabatic process is one for which no heat flows into or out of a system. For an
Example:
P (kPa)
An ideal monatomic gas goes through the
cyclic process ABCA as shown to
the right. The temperature of the gas at A is
600K. Calculate the work done on the gas,
the heat absorbed by the gas, and the change
in internal energy for each process and for
the total cycle.
30
A
20
10
B
C
0
From the ideal gas equation, PV = nRT, we
0
2
can calculate that TB = 600K and TC =
200K. Since this is a monatomic gas, then
we have U = 3/2 nRT. We keep in mind that 1 liter = 10-3 m3.
V (L)
4
6
AB:
Since the gas is expanding, then the work done on the gas is negative and
WAB = area under PV curve = (2x104 Pa)(4x10-3 m3) = 80 J
UAB = 3/2 nRTAB = 0 (TA = TB)
From the 1st law, QAB = UAB + WAB = 0 + 80 J = 80 J (heat is absorbed)
2
BC:
Since the gas is being compressed, the work done on the gas is negative and
since the area under PV curve = (1x104 Pa)(4x10-3 m3) = 40 J
WBC = -40J
UBC = 3/2 nRTBC = 3/2 nR(TC – TB) = 3/2 PCVC – 3/2 PBVB
=3/2 (1x104 Pa)(2 – 6 )x10-3 m3 = -60 J
QBC = UBC + WBC = -60 J+ (– 40) J = -100 J
CA:
This is an isovolumeric process, so WCA = 0.
UCA = 3/2 nRTCA = 3/2 nR(TA – TC) = 3/2 PAVA – 3/2 PCVC
=3/2 (3x104 Pa – 1x104 Pa)(2x10-3 m3) = 60 J
QCA = UCA + WCA = 60 J
Summary results: U = QNET - WNET
AB BC CA NET
W (J)
+80
-40
0
40
Q (J)
80
-100
60
40
0
-60
60
0
U (J)
Note that the total work done in the cycle is
the enclosed area. The negative value means
that the gas does work on its environment
during the cycle. Since the gas returns to its
original state, the net change in internal
energy is zero. The net heat absorbed is equal
to the net work done by the gas. Heat is
absorbed during the processes AB and
CA and rejected during the process BC.
P (kPa)
30 Qin
area =
work
20
10
B
C
Qout
0
V (L)
0
3
Qin
A
2
4
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Heat Engines and the 2nd Law of Thermodynamics
The gas undergoing the cyclic process described in the above problem is an example of a
heat engine. It absorbs heat at high temperatures, dumps heat at low temperatures, and
converts the difference into work.
Qin
P
Hot
Area =
work
Qin
Qout
W
Qout
V
Cold
The efficiency of a heat engine is the ratio of the net work done during the cycle to the
heat absorbed.
Since ΔU=0 it follows QNET = WNET with QNET = Qin + Qout = Qin - |Qout|. The efficiency
of the heat engine is given by
e
W
Qin
Where W is do work done by the system = WNET
e
Qin  Qout
Qin
 1
| Qout |
Qin
The 2nd law of thermodynamics states that no heat engine can have an efficiency that is
100% (e = 1). In other words, a heat engine cannot extract heat from a reservoir and
convert it completely to work. Some heat must be dumped at lower temperatures.
Example:
In the previous example of the ideal monatomic gas undergoing a cyclic process,
calculate the efficiency.
Qin  Q AB  QCA  80 J  60 J  140 J
W (net )  40 J (done b y the gas )
e
W
40 J

 0.286  28.6%
Qin 140 J
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Maximum possible efficiency (Carnot heat engine)
The maximum possible efficiency of a heat engine that absorbs heat at Thot and dumps
heat at Tcold is
e  1
Tc
Th
This would apply if the heat engine were an ideal gas. All other heat engines (Diesel,
Otto, etc...) with ideal gases have a lower efficiency than the Carnot cycle.
Example:
A heat engine absorbs heat at 500oC and dumps heat at 25oC. What is the maximum
efficiency?
e  1
Tc
25  273
 1
 0.386
Th
500  273
If the engine takes in heat at the rate of 10 kW, what is the power output?
W  e Qin
Power 
Qin
W
e
 (0.368)(10kW )  3.68kW
t
t
In reality, since friction would be involved, more heat is actually lost by the heat engine
(at each transformation) and the real efficiency is less:
e real engine &lt; e ideal engine
Entropy
The entropy of a system is a measure of its disorder. The higher the disorder, the higher
the entropy. Specifically for a system which absorbs heat at a fixed temperature, then the
change in entropy is given by
S 
Q
T
units = J/K or cal/K
If heat is absorbed, then S &gt; 0. If heat is lost, then S &lt; 0.
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Example:
50 g of water melts at 0oC. What is the change in entropy of the water?
S 
Q mL f
(50 g )(80cal / g )


 14.7 cal / K  61.3 J / K
T
T
273K
Example:
Heat of 100cal is transferred from an object at 100&deg;C to another at 0&deg;C. Find the total
change in entropy.
Qhot = -100 cal
Qcold = 100cal
negative sign since this heat is released
positive sign since this heat is absorbed
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Example:
100 cal of heat is transferred from a reservoir at 100oC to a reservoir at 0oC. Assuming
that the reservoirs are large enough so that their temperatures don’t change, what is the
total change in entropy of the two reservoirs?
S  S hot  S cold 
Qhot Qcold  100cal 100cal



Thot Tcold
373K
273K
 0.098 cal / K
Note that S &gt; 0, which means that the total disorder has increased. If heat flowed from
the cold to the hot, then S would be negative. This cannot occur spontaneously.
Another way of stating the 2nd law of thermodynamics is to say that the total entropy of
a closed system increases in all natural processes.
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