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601_PartUniversity Physics Solution

# 601_PartUniversity Physics Solution - 20-16 Chapter 20...

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20-16 Chapter 20 20.49. IDENTIFY: Use UQW Δ=− and the appropriate expressions for Q , W and U Δ for each type of process. p Vn R T = relates T Δ to p and V values. H , W e Q = where H Q is the heat that enters the gas during the cycle. SET UP: For a monatomic ideal gas, 53 22 and C . PV CR R == (a) ab : The temperature changes by the same factor as the volume, and so 5 ( ) (2.5)(3.00 10 Pa)(0.300 m ) 2.25 10 J. P Pa a b C Qn CT pV V R = −= × = × The work p V Δ is the same except for the factor of 5 5 2 , so 0.90 10 J. W 5 1.35 10 J. Δ=− = × bc : The temperature now changes in proportion to the pressure change, and 3 53 5 2 ( ) (1.5)( 2.00 10 Pa)(0.800 m ) 2.40 10 J, cb b Qp p V =− × = × and the work is zero 5 ( 0). 2.40 10 J. VU Q W Δ= Δ= − = × ca : The easiest way to do this is to find the work done first; W will be the negative of area in the p - V plane bounded by the line representing the process ca and the verticals from points a and c . The area of this trapezoid is 55 3 3 4 1 2 (3.00 10 Pa 1.00 10 Pa)(0.800 m 0.500 m ) 6.00 10 J ×+ × = × and so the work is 5 0.60 10 J. −× U Δ must be 5 1.05 10 J (since 0 U ×Δ = for the cycle, anticipating part (b)), and so Q must be 5 0.45 10 J. UW Δ+ = × (b) See above; 5 0.30 10 J, 0. QW U == × Δ= (c) The heat added, during process ab and ca , is 2.25 10 J 0.45 10 J × 5 2.70 10 J and the efficiency is 5 5 H 0.30 10 0.111 11.1%. 2.70 10 W e Q × = = × EVALUATE: For any cycle, 0 U and . = 20.50. IDENTIFY: Use the appropriate expressions for Q , W and U Δ for each process. H / eWQ = and Carnot C H 1/ . eT T SET UP: For this cycle, H2 TT = and C1 = EXECUTE: (a) ab : For the isothermal process, 0 T Δ = and 0. U Δ = 11 1 ln( ) ln(1/ ) ln( ) ba Wn R T VV n R T r n R Tr = and 1 ln( ). QW n R T r bc : For the isochoric process, 0 V and 0. W = 21 () . VV QU n C T n C T T = Δ = cd : As in the process ab , 2 0 and ln( ). Q n R T r da : As in process bc , 0 and 0; VW = 12 . V UQn CTT Δ== (b) The values of Q for the processes are the negatives of each other. (c) The net work for one cycle is net 2 1 l n ( ) , R T and the heat added (neglecting the heat exchanged during the isochoric expansion and compression, as mentioned in part (b)) is cd 2 ln( ), R = and the efficiency is net 1( ) . cd W T Q This is the same as the efficiency of a Carnot-cycle engine operating between the two temperatures. EVALUATE: For a Carnot cycle two steps in the cycle are isothermal and two are adiabatic and all the heat flow occurs in the isothermal processes. For the Stirling cycle all the heat flow is also in the isothermal steps, since the net heat flow in the two constant volume steps is zero. 20.51. IDENTIFY: The efficiency of the composite engine is 12 H1 , WW e Q + = where H1 Q is the heat input to the first engine and 1 W and 2 W are the work outputs of the two engines. For any heat engine, CH , WQ Q =+ and for a Carnot engine, low low high high , QT where low Q and high Q are the heat flows at the two reservoirs that have temperatures low T and high . T SET UP: high,2 low,1 .

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601_PartUniversity Physics Solution - 20-16 Chapter 20...

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