A 1.00-mol sample of a monatomic ideal gas is taken through the cycle shown in Figure P22.76. At point A, the pressure, volume, and temperature are Pi, Vi, and Ti, respectively. In terms of R and Ti, find (a) the total energy entering the system by heat per cycle, (b) the total energy leaving the system by heat per cycle, and (c) the efficiency of an engine operating in this cycle. (d) Explain how the efficiency compares with that of an engine operating in a Carnot cycle between the same temperature extremes.
(a)
The total energy entering the system by heat per cycle.
Answer to Problem 22.76AP
The total energy entering the system by heat per cycle is
Explanation of Solution
For work done from point
The work done is,
The ideal gas equation for temperature at point
Here,
Consider the point
Substitute
The ideal gas equation for temperature at point
Here,
Substitute
The heat equation between point
Here,
Substitute
For work done from point
The work done is,
Here,
Substitute
The formula for the internal energy is,
Here,
The ideal gas equation for temperature at point
Here,
Substitute
Substitute
Substitute
The energy transferred is,
Substitute
For work done from point
The work done is,
The ideal gas equation for temperature at point
Here,
Substitute
The energy transferred is,
Substitute
Substitute
For work done from point
The work done is,
Substitute
The energy transferred as heat is,
Here,
Substitute
Substitute
The net energy entering the system is,
Substitute
Conclusion:
Therefore, the total energy entering the system by heat per cycle is
(b)
The total energy leaving the system by heat per cycle.
Answer to Problem 22.76AP
The total energy leaving the system by heat per cycle is
Explanation of Solution
The total energy leaving the system is,
Substitute
Conclusion:
Therefore, the total energy leaving the system by heat per cycle is
(c)
The efficiency of an engine operating in this cycle.
Answer to Problem 22.76AP
The efficiency of an engine operating in this cycle is
Explanation of Solution
The formula of efficiency of the engine is,
Conclusion:
Therefore, the efficiency of an engine operating in this cycle is
(d)
The efficiency comparison with that of an engine operating in a Carnot cycle between the same temperature extremes.
Answer to Problem 22.76AP
The efficiency of an engine is
Explanation of Solution
The maximum temperature in cycle is,
The minimum temperature in cycle is,
The formula of the Carnot efficiency is,
Substitute
Conclusion:
Therefore, the efficiency of an engine is
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Chapter 22 Solutions
Physics for Scientists and Engineers, Technology Update (No access codes included)
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