Physics 1308 Quiz Chapter 14 SOLUTION
Physics 1308 Quiz Chapter 14 SOLUTION Problem A reversible engine contains 0.2mol of ideal monoatomic gas (). The gas undergoes the cycle below (the curved segment represents an isothermal, P
6.0 atm, 2.0 L A
C 2.0 atm 2.0 L
B 2.0 atm 6.0 L
V (a) Determine the net work done . (b) Determine the efficiency of the cycle. (You need to find all the heat absorbed) Solution (a) The net work done is equal to the area enclosed in the cycle. This area is equal to the area below the isothermal minus the area below the isobaric process (There is no work performed between A and C). Thus: V W net =W A BW B C =nRT A B ln B P V B C VA We need first the temperature of the isothermal. Using PV=nRT we obtain : 6.0 atm×2 L T A B = =732 K L atm 0.20mol×0.082 mol K we are ready. Therefore, L atm W net =0.20mol∗0.082 732 K ln 32.0 atm 2.0L−6.0 L =5.2 L atm mol K 1/2
(b) In order to determine the efficiency we need to find the total heat absorbed. Observing the cycle we can deduce the following: From A>B the gas expands at constant temperature, thus it has to absorb heat in order to compensate for the work done, since there is no change in the internal energy. From B>C the gas is compressed and it actually lowers its temperature, thus it has to expel heat. From C>A the gas increases temperature without any work involved, then it has to absorb heat. So, Q h =Q A B Q C A lets us find each one of them. Since A>B is isothermal, the work done is equal to the heat absorbed: Q A B =W A B=13.2 L atm From C>A we know that the change in internal energy is equal to the heat absorbed (zero work here!) therefore we need to find the temperature of the gas at point C. Once more: 2.0 atm×2 L T C= =244 K L atm 0.20mol×0.082 mol K and the change of internal energy is given by: 3 3 L atm U= n R T = 0.20mol 0.082 732K−244K=12 L atm=QC A 2 2 mol K with all pieces into place we can evaluate the efficiency: W 5.2 L atm e= = =0.21 Q h 13.2 L atm12 L atm Just for fun we can compare this efficiency with the ideal: T 244K e ideal =1− l =1− =0.67 Th 732K which gives us an idea of how bad the cycle is.
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6.0 atm, 2.0 L A
C 2.0 atm 2.0 L
B 2.0 atm 6.0 L
V (a) Determine the net work done . (b) Determine the efficiency of the cycle. (You need to find all the heat absorbed) Solution (a) The net work done is equal to the area enclosed in the cycle. This area is equal to the area below the isothermal minus the area below the isobaric process (There is no work performed between A and C). Thus: V W net =W A BW B C =nRT A B ln B P V B C VA We need first the temperature of the isothermal. Using PV=nRT we obtain : 6.0 atm×2 L T A B = =732 K L atm 0.20mol×0.082 mol K we are ready. Therefore, L atm W net =0.20mol∗0.082 732 K ln 32.0 atm 2.0L−6.0 L =5.2 L atm mol K 1/2
(b) In order to determine the efficiency we need to find the total heat absorbed. Observing the cycle we can deduce the following: From A>B the gas expands at constant temperature, thus it has to absorb heat in order to compensate for the work done, since there is no change in the internal energy. From B>C the gas is compressed and it actually lowers its temperature, thus it has to expel heat. From C>A the gas increases temperature without any work involved, then it has to absorb heat. So, Q h =Q A B Q C A lets us find each one of them. Since A>B is isothermal, the work done is equal to the heat absorbed: Q A B =W A B=13.2 L atm From C>A we know that the change in internal energy is equal to the heat absorbed (zero work here!) therefore we need to find the temperature of the gas at point C. Once more: 2.0 atm×2 L T C= =244 K L atm 0.20mol×0.082 mol K and the change of internal energy is given by: 3 3 L atm U= n R T = 0.20mol 0.082 732K−244K=12 L atm=QC A 2 2 mol K with all pieces into place we can evaluate the efficiency: W 5.2 L atm e= = =0.21 Q h 13.2 L atm12 L atm Just for fun we can compare this efficiency with the ideal: T 244K e ideal =1− l =1− =0.67 Th 732K which gives us an idea of how bad the cycle is.
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