Physics 1308 Practice Quiz Chapter 14
Physics 1308 Practice Quiz Chapter 14
Problem A 0.20mol sample of an ideal gas goes through the Carnot cycle of the figure below. Calculate (a) the heat Qh absorbed, (b) the heat Ql rejected, and (c) the work done. (d) Use these quantities to determine the efficiency. (e) Find the maximum and minimum temperatures and show explicitly that the efficiency is the ideal efficiency. P 8.0 atm , 1 L A. B. D. 4.1 atm, 1.612 L
4.0 atm, 2.0 L
C. 2.050 atm, 3.224 L
V Problem Solving 1. Define the system you are dealing with; be careful to distinguish the system under study from its surroundings. 2. Be careful of signs associated with work and heat. In the first law work done by the system is positive; work done on the system is negative. Heat added to the system is positive but heat removed from the system it is negative. With heat engines, we usually consider heat and work as positive and write energy conservation equations with + and – signs taking into account directions. 3. Watch the units used for work and heat. 4. Temperatures must generally be expressed in kelvins. 5. Efficiency (or coefficient of performance) is a ratio of two energies or powers: useful output divided by required input. 6. The entropy of a system increases when heat is added to the system, and decreases when heat is removed. If heat is transferred from system A to system B, the change in entropy of A is negative and the change of entropy of B is positive. 1/3
Solutions The Carnot cycle needs two different temperatures to operate. Let's find them. Since the gas is an ideal gas we can use PV=nRT for both isothermals to obtain : 8.0 atm×1 L T h= =390 K L atm 0.25mol×0.082 mol K 4.1 atm×1.612 L T l= =322 K L atm 0.25mol×0.082 mol K These are the operating temperatures of the cycle. Now let us analyze each step of the loop. A>B: Isothermal expansion Using the first law we have U=Q−W 0=Q−W Q=W this relation together with the equation for work in an isothermal process V W A B =nRT h ln B VA
2.0 L L atm ×390K×ln mol K 1.0 L W A B =5.6 L atm
W A B=0.25mol×0.082
allow us to find: Qh =5.6 L atm .
B>C: Adiabatic expansion (from Th to Tl ) Again using the first law we get U=Q−W U =0−W U=−W Now we need the equation that relates the change in internal energy with the change in temperatures:
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3 U = n R T 2 3 L atm U= ×0.25mol×0.082 ×322K −390 2 mol K U=−2.1 L atm so the work done by the system is 2.1 L atm. Note that the system is expanding, performing work on the environment. This fact explains the positive sign. C>D: Isothermal compression This process is similar to A>B, but with compression instead of expansion. Therefore we get: W C D=nRT l ln
VD VC
1.612 L L atm ×322K×ln mol K 3.224 L W A B=−4.6 L atm
W A B=0.25mol×0.082
The negative value means that work is done on the gas and to maintain the temperature constant it is necessary to expel heat in the same amount. This heat is equal to Ql . D>A: Adiabatic compression (from Tl to Th ) The value here for the change in internal energy is easily found since now we are reversing the process from B>C thus: U=2.1 L atm and the work done is 2.1 L atm. We are now ready to find the net work : W net =5.62.1−4.6−2.1 L atm=1 L atm We could have obtained the same result using: W net =Qh −Ql=5.6−4.6 L atm=1 L atm . The efficiency of the cycle is: W 1 L atm e= = =0.18 Qh 5.6 L atm
using the Carnot efficiency we find: T 322K e ideal =1− l =1− =0.18 Th 390K The equality between both efficiencies confirms our calculations.
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Problem A 0.20mol sample of an ideal gas goes through the Carnot cycle of the figure below. Calculate (a) the heat Qh absorbed, (b) the heat Ql rejected, and (c) the work done. (d) Use these quantities to determine the efficiency. (e) Find the maximum and minimum temperatures and show explicitly that the efficiency is the ideal efficiency. P 8.0 atm , 1 L A. B. D. 4.1 atm, 1.612 L
4.0 atm, 2.0 L
C. 2.050 atm, 3.224 L
V Problem Solving 1. Define the system you are dealing with; be careful to distinguish the system under study from its surroundings. 2. Be careful of signs associated with work and heat. In the first law work done by the system is positive; work done on the system is negative. Heat added to the system is positive but heat removed from the system it is negative. With heat engines, we usually consider heat and work as positive and write energy conservation equations with + and – signs taking into account directions. 3. Watch the units used for work and heat. 4. Temperatures must generally be expressed in kelvins. 5. Efficiency (or coefficient of performance) is a ratio of two energies or powers: useful output divided by required input. 6. The entropy of a system increases when heat is added to the system, and decreases when heat is removed. If heat is transferred from system A to system B, the change in entropy of A is negative and the change of entropy of B is positive. 1/3
Solutions The Carnot cycle needs two different temperatures to operate. Let's find them. Since the gas is an ideal gas we can use PV=nRT for both isothermals to obtain : 8.0 atm×1 L T h= =390 K L atm 0.25mol×0.082 mol K 4.1 atm×1.612 L T l= =322 K L atm 0.25mol×0.082 mol K These are the operating temperatures of the cycle. Now let us analyze each step of the loop. A>B: Isothermal expansion Using the first law we have U=Q−W 0=Q−W Q=W this relation together with the equation for work in an isothermal process V W A B =nRT h ln B VA
2.0 L L atm ×390K×ln mol K 1.0 L W A B =5.6 L atm
W A B=0.25mol×0.082
allow us to find: Qh =5.6 L atm .
B>C: Adiabatic expansion (from Th to Tl ) Again using the first law we get U=Q−W U =0−W U=−W Now we need the equation that relates the change in internal energy with the change in temperatures:
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3 U = n R T 2 3 L atm U= ×0.25mol×0.082 ×322K −390 2 mol K U=−2.1 L atm so the work done by the system is 2.1 L atm. Note that the system is expanding, performing work on the environment. This fact explains the positive sign. C>D: Isothermal compression This process is similar to A>B, but with compression instead of expansion. Therefore we get: W C D=nRT l ln
VD VC
1.612 L L atm ×322K×ln mol K 3.224 L W A B=−4.6 L atm
W A B=0.25mol×0.082
The negative value means that work is done on the gas and to maintain the temperature constant it is necessary to expel heat in the same amount. This heat is equal to Ql . D>A: Adiabatic compression (from Tl to Th ) The value here for the change in internal energy is easily found since now we are reversing the process from B>C thus: U=2.1 L atm and the work done is 2.1 L atm. We are now ready to find the net work : W net =5.62.1−4.6−2.1 L atm=1 L atm We could have obtained the same result using: W net =Qh −Ql=5.6−4.6 L atm=1 L atm . The efficiency of the cycle is: W 1 L atm e= = =0.18 Qh 5.6 L atm
using the Carnot efficiency we find: T 322K e ideal =1− l =1− =0.18 Th 390K The equality between both efficiencies confirms our calculations.
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