The Thermodynamic Temperature and the Zeroth Law of Thermodynamics
Section 3.1: The Second Law of Thermodynamics and the Carnot Heat Engine
3.1 a. A Carnot engine contains 0.250 mol of a monatomic ideal gas as its working fluid. AssumeCVto be constant and equal to 3nR/2. IfTh473 K,Tc373 K, V10.600 L, and if the compression ratio (the ratio V3/V1) equals 6.00, find the efficiency and the values of V2andV4.
b. Calculatewfor each of the steps in the cycle of part a.
3.2 A Carnot engine contains as working fluid 0.150 mol of neon (assume ideal) withCV,m3R/2. IfTh600 K and Tc373 K, find the values ofV1,V2,V3,P3,P4, andV4if P120.00 atm andP25.00 atm.
3.3 Carbon monoxide is used as the fuel for a Carnot engine with a high temperature of 450◦C and a cool temperature of 100◦C. Determine how high the combustion of 1.000 mol of CO could lift a 1.00 kg mass near the surface of the earth. Assume that all of the heat from the combustion is transferred to the engine and assume that the combustion takes place at 450◦C.
3.4 A Carnot engine is operated by burning 1.00 mol of methane at a temperature of 1000 K. The cool reservoir is at 373 K.
a. Find the enthalpy change of the combustion reaction at 1000 K.
b. Find the amount of work that can be done by the engine when it burns this much fuel. Assume that no heat is lost.
3.5 a. A steam engine operates with its boiler at 200◦C and a pressure of 15.34 atm, and with its exhaust at a temperature of 100◦C and a pressure of 1.000 atm. Find the Carnot efficiency for these temperatures. What can you say about the efficiency of the steam engine?
b. The boiler is reinforced to operate at 360◦C and a pres- sure of 184 atm. If the exhaust remains at 100◦C, find the percentage improvement in the Carnot efficiency.
c. If the coal that the steam engine in part b burns is assumed to be pure graphite (not a good assumption) find the mass of coal required to produce 10.00 horsepower for 1.000 hour, assuming the Carnot efficiency. 1 horsepower746 watt746 J s−1. 3.6 Assume gasoline to be puren-octane and assume that its
enthalpy change of combustion is approximately
independent of temperature. Calculate the height to which combustion of 1.000 kg of gasoline can lift an automobile with a mass of 1500 kg. Assume that the automobile engine has a combustion temperature of 2000◦F and an exhaust temperature of 1100◦F and that the engine has an efficient 75% as large as that of a Carnot engine operating between these temperatures.
3.7 a. Assume that natural gas is pure methane (it is actually 90–95% methane). Find the amount of heat put into a house by the combustion of 100 cubic feet of natural gas at 20.0◦C and 1.00 atm if 20.0% of the heat is wasted (hot gases go up the flue).
b. Find the amount of heat put into a house if 100.0 cubic feet of natural gas at 20.0◦C and 1.00 atm is burned in an electric generating plant that is 80.0% as efficient as a Carnot engine operating between 2000◦C and 800◦C and if 90% of this energy is delivered to a heat pump operating between 20.0◦C and 0.00◦C. Assume that this heat pump has a coefficient of performance that is 80.0% of that of a Carnot heat pump.
c. Find the amount of heat put into a house if 100.0 cubic feet of natural gas is burned in an electric generating plant that is 80.0% as efficient as a Carnot engine operating between 2000◦C and 800◦C and if 90% of this energy is delivered to a resistance heater that has 100% efficiency.
3.8 a. Calculate the coefficient of performance of a Carnot food freezer with an interior temperature of−18◦C and an exterior temperature of 25◦C. 1 watt1 J s−1. b. Calculate the amount of electrical energy in
kilowatt-hours necessary to freeze 1.000 kg of water in the Carnot freezer of part a.
3.9 A reversible heat engine accepts heat from a hot reservoir at temperatureT1but exhausts part of the heat (−q2) at
temperatureT2and part (−q3) of it at temperatureT3, whereT2> T3. Find the efficiency ifq3 T2
T3
q2.
3.10 a. Assume that the human body has the same efficiency in doing mechanical work as a Carnot engine with an upper temperature equal to human body temperature, 37◦C, and a lower temperature equal to 25◦C. Find the efficiency.
b. The actual efficiency of the human body is
approximately equal to 20%. How can you explain this, since no heat engine can be more efficient than a Carnot engine?
3.11 It has been proposed that a heat engine might economically operate using the temperature difference between sea water near the surface and at a depth of several hundred feet.
a. Assume that such a heat engine has 50% of the efficiency of a Carnot engine and operates between 30◦C and 20◦C. Find the efficiency of the
engine.
b. Assume that the heat engine drives an electric generator that produces 100 Mwatt (100 megawatts). Find the volume of sea water that must pass through the high-temperature heat exchanger per second if the heat exchanger lowers the temperature of the sea water from 30◦C to 20◦C. Assume that the sea water has the same heat capacity as pure water at 298.15 K, 75.351 J K−1mol−1, and density equal to 1.00×103kg m−3.
3.2 The Mathematical Statement of the Second Law: Entropy
The second law of thermodynamics can be stated mathematically in a way that defines a new state function:If the differentialdSis defined by
dS dqrev
T (definition of the entropyS) (3.2-1) thendSis an exact differential andSis a state function called the entropy.We now show thatdqrev/Tis an exact differential. The integral of an exact differential is path- independent, so that its integral around a closed path (a path that starts and ends at the same point) vanishes. The converse is also true: If the integral around an arbitrary closed path vanishes, the differential is exact. We need to show that
dqrev
Tsurr 0 (3.2-2)
for all reversible cyclic processes in a closed system. The symbol
represents a line integral around a closed curve in the state space (beginning and ending at the same state).
We begin with a Carnot cycle. From Eqs. (3.1-13) and (3.1-23) q1
Th −q3
Tc
(3.2-3) SinceT is constant on the isothermal segments and sincedqrev0 on the adiabatic segments, the line integral for a Carnot cycle is
◦ dqrev
T 1 Th
V2
V1
dqrev+0+ 1 Tc
V4
V3
dqrev+0 q1
Th +q3
Tc 0 (3.2-4) so that Eq. (3.2-2) is established for any Carnot cycle.
We now show that Eq. (3.2-2) is valid for the reversible cyclic process of Figure 3.4a. Steps 1, 3, and 5 are isothermal steps, and steps 2, 4, and 6 are adia- batic steps. Let point 7 lie on the curve from state 6 to state 1, at the same temperature as states 3 and 4, as shown in Figure 3.4b. We now carry out the reversible cyclic process 1→2→3→7→1, which is a Carnot cycle and for which the line integral vanishes. We next carry out the cycle 7→4→5→6→7. This is also a Carnot cycle, so the line integral around this cycle vanishes. During the second cycle the path from state 7 to state 3 was traversed from left to right. During the first cycle, the path from state 3 to state 7 was traversed from right to left. When the two cyclic line integrals are added, the integrals on these two paths cancel each other, and if we leave them both out the sum of the two line integrals is unchanged. The sum of the two line integrals is now a vanishing line integral for the cyclic process 1→2→3→4→5→6→ 7→1, which is the cycle of Figure 3.4a.
We now show that Eq. (3.2-2) holds for any cyclic process made up of isothermal and adiabatic reversible steps. Consider the process of Figure 3.5a, which can be divided into three Carnot cycles, just as that of Figure 3.4a was divided into two Carnot cycles.
We can do the same division into Carnot cycles for any cycle that is made up of reversible isothermal and adiabatic steps. If each Carnot cycle is traversed once, the integrals on all of the paths in the interior of the original cycle are traversed twice, once in each direction, and therefore cancel out when all of the line integrals are added together. The exterior curve is traversed once, and the integral of Eq. (3.2-2) is shown to vanish around the cycle. For example, Figure 3.5b shows a cycle equivalent to eight
V
6 6
7
1 2
3 4
5
1 2
3 4
5
(a)
T
V (b)
T
Figure 3.4 A Reversible Cycle of Isotherms and Adiabats. (a) The original cycle.
(b) The cycle with an added process.
V
V
T
(a)
(c)
T
V (b)
T
Figure 3.5 Reversible Cycles of Isotherms and Adiabats.(a) A cycle equivalent to three Carnot cycles. (b) A cycle equivalent to eight Carnot cycles. (c) A cycle equivalent to a large number of Carnot cycles.
Carnot cycles, and Figure 3.5c shows a more complicated cycle that can be divided into a large number of Carnot cycles. We conclude that the line integral of dqrev/T around any path consisting of reversible isotherms and adiabats vanishes.
In order to represent an arbitrary cycle we construct reversible isothermal and adi- abatic steps that are smaller and smaller in size, until the curve of the arbitrary cycle is more and more closely approximated by isothermal and adiabatic steps. In the limit that the sizes of the steps approach zero, any curve is exactly represented and the line integral ofdqrev/T vanishes for any cycle. The differentialdSdqrev/T is therefore exact andSis a state function. For a simple system containing one substance and one phase,Smust be a function of three state variables. We can write
SS(T,V,n) (3.2-5)
or
SS(U,V,n) (3.2-6)
and so on. Because we have defined only the differential of the entropy, any constant can be added to the value of the entropy without any significant change, just as a constant can be added to the value ofUwithout any significant change.
Caratheodory1 showed that dqrev/T is an exact differential by a more formal mathematical procedure. His argument is sketched briefly in Appendix D. It begins with the fact that two reversible adiabats cannot cross. We now show that this is a fact. We have already seen an example of it in the previous chapter when we derived a formula for the reversible adiabat for an ideal gas with a constant heat capacity.
Equation (3.4-21b) is
T T1
V1
V nR/CV
(3.2-7) This equation represents a family of functions ofT as a function ofV, one for each initial state:
T T(V) (3.2-8)
No two curves in this family can intersect.
Constantin Caratheodory, 1872–1950, was a Greek-German mathematician who made many contributions in mathematics in addition to his work in thermodynamics.
To show that two reversible adiabats cannot cross for other systems we assume the opposite of what we want to prove and then show that this assumption leads to a contradiction with fact and therefore must be false. Assume that there are two different reversible adiabats in the state space of a closed simple system and that the curves coin- cide at state number 1, as depicted in Figure 3.6. We choose a state on each reversible adiabat, labeled state number 2 and state number 3 such that the reversible process leading from state 2 to state 3 hasq >0. Now consider a reversible cyclic process 1→2→3→1. Since steps 1 and 3 are adiabatic,
qcycleq2>0 (3.2-9)
Since∆U0 in any cyclic process,
wsurr −wcycleqcycleq2 (3.2-10)
Heat transferred to a system undergoing a cyclic process has been converted completely to work done on the surroundings, violating the second law of thermodynamics. The source of this violation is the assumption that two reversible adiabats can cross. There- fore, only one reversible adiabat passes through any given state point. The rest of Caratheodory’s argument is summarized in Appendix D.
V
T
Two proposed adiabats that cross 2
3
1
Figure 3.6 Two Reversible Adiabats That Cross (Assumption to Be Proved False).