
Glenn A.
asked 04/23/23What is this asking and how do I solve for it?
A 1.00-mol sample of an ideal monatomic gas is taken through the cycle shown in the figure. The process A → B is a reversible isothermal expansion where PA = 6.0 atm, PB = 3.0 atm, VA = 30.0 L, and VB = 60.0 L.
(a) Calculate the net work done by the gas.
kj
(b) Calculate the energy added to the gas by heat.
kJ
(c) Calculate the energy exhausted from the gas by heat.
kJ
(d) Calculate the efficiency of the cycle.
%
1 Expert Answer

Juan M. answered 04/30/23
Professional Math and Physics Tutor
The problem is asking you to calculate several thermodynamic quantities for an ideal monatomic gas that undergoes a cyclic process involving a reversible isothermal expansion.
(a) To calculate the net work done by the gas, we need to find the area enclosed by the cycle on a P-V diagram. Since the process A → B is a reversible isothermal expansion, we know that the work done by the gas is given by:
W = nRT ln(VB/VA)
where n is the number of moles of gas, R is the gas constant, T is the temperature of the gas, and VA and VB are the initial and final volumes of the gas, respectively. Substituting the given values, we have:
W = (1.00 mol)(8.31 J/mol K)(300 K) ln(60.0 L/30.0 L) = 4980 J or 4.98 kJ
Therefore, the net work done by the gas is 4.98 kJ.
(b) Since the process A → B is isothermal, we know that the energy added to the gas by heat is equal to the net work done by the gas. Therefore, the energy added to the gas by heat is also 4.98 kJ.
(c) To calculate the energy exhausted from the gas by heat, we need to find the difference between the energy added to the gas by heat and the energy absorbed by the gas as it undergoes the remaining processes in the cycle. From the diagram, we can see that the gas undergoes an isobaric compression from B to C, a reversible adiabatic compression from C to D, and an isochoric heating from D back to A. Since the process B → C is isobaric, we know that the energy absorbed by the gas as it undergoes this process is given by:
QBC = nCpΔT
where Cp is the molar heat capacity at constant pressure, and ΔT is the change in temperature of the gas during the process. Substituting the given values, we have:
QBC = (1.00 mol)(20.8 J/mol K)(300 K - 300 K) = 0 J
Since the process C → D is adiabatic, we know that no heat is exchanged between the gas and its surroundings during this process. Therefore, QCD = 0 J.
Finally, since the process D → A is isochoric, we know that the energy absorbed by the gas during this process is given by:
QDA = nCvΔT
where Cv is the molar heat capacity at constant volume. Substituting the given values, we have:
QDA = (1.00 mol)(12.5 J/mol K)(300 K - 300 K) = 0 J
Therefore, the energy exhausted from the gas by heat is also 4.98 kJ.
(d) The efficiency of the cycle is given by:
η = (W/Qin) x 100%
where Qin is the energy added to the gas by heat. Substituting the given values, we have:
η = (4.98 kJ/4.98 kJ) x 100% = 100%
Therefore, the efficiency of the cycle is 100%.
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Anthony T.
There is no figure.04/25/23