# Chapter 12 * The Laws of Thermodynamics

```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 by the gas is positive and
WAB = area under PV curve = 1/2(2x104 Pa)(4x10-3 m3)+ (104 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, work is done on the gas. The change in
volume ΔV is negative, and so the work given by the area under PV curve is
WBC = PΔV= = (1x104 Pa)(-4x10-3 m3) = -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 (heat is released)
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 net work done in the cycle is the
enclosed area. The positive 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, in the case of heat engine, 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: Qout cannot be zero.
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 found to be
e  1
Tc
Th
This would apply if the heat engine were an ideal gas. All other heat engines (Diesel,
Otto, etc...) working using 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 is present, more heat is actually lost by the heat engine (at each
transformation) and the real efficiency is less than the ideal engine’s efficiency:
e real engine &lt; e ideal engine
Entropy
Mathematically, entropy is a function of the state of a system: it characterizes the system
equivalently to its volume, mass or temperature. As for the potential energy, only changes
in entropy are significant. For the case of a system which exchanges heat at a fixed
temperature, the change in entropy is
S 
Q
T
units = J/K or cal/K
If heat is absorbed, then Q &gt; 0. If heat is lost, then Q &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
Since Q &gt; 0 for this process ΔS &gt; 0
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?
Qhot = -100 cal
Qcold = 100cal
negative sign since this heat is released
positive sign since this heat is absorbed
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 entropy 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.
∆𝑆 ≥ 0
Where ΔS = 0 corresponds to a reversible ideal process (with no friction).
A more general definition of entropy is found in statistical mechanics where entropy is
related to the amount of ‘disorder’ of an isolated system: the higher the disorder, the
higher the entropy. An absolute value (not just the ΔS) of entropy can be evaluated:
𝑆 = 𝑘𝐵 ln 𝑊
where ln is the natural logarithm and W is the number of microstates.
The value of W indicates the number of possibilities for a macroscopic configuration M
to occur. The higher is W the most likely that configuration occurs.
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Example:
Consider tossing a coin four times, the possible outcomes (the macroscopic configuration
M) are:
M
4 tails
HHHH
HHHT, HHTH, HTHH, THHH
HHTT, HTHT, HTTH, THHT, THTH, TTHH
HTTT, HHTH, HTHH, THHH
TTTT
W
1
4
6
4
1
For the case M = two heads and two tails W = 6 which is the highest value for W. That
means W = 6 is the configuration most favorable to occur when tossing a coin four times.
It also corresponds to the highest entropy.
The 2nd law of thermodynamics now read: processes in nature occur in such a way that
the disorder of the system naturally increases, such systems spontaneously evolve
towards thermodynamic equilibrium, the configuration with maximum entropy.
Example:
Put inside a container one cup of salt, then add one cup of pepper, shake the container.
The macroscopic configuration most favorable to occur is the mixture between salt and
pepper, you can keep shaking the container but it’s very unlikely to obtain the initial
configuration again. For the initial macroscopic configuration the value of W is very
small, while the value of W for the mixture is large since it can be realized in many
different ways.
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