Chapter 20, Problem 033 Your answer is partially correct. Try again. The figure
ID: 1869304 • Letter: C
Question
Chapter 20, Problem 033 Your answer is partially correct. Try again. The figure shows a reversible cycle through which 1.00 mole of a monatomic ideal gas is taken. Process bc is an adiabatic expansion, with Po 10.0 atm and Vo = 2.30 x 10 3 m. For the cycle find (a) the energy added to the gas as heat, (b) the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle. Adiabatic 8.00V Volume (a) NumberT (b)Number (c) Number (d) NumberT Units nober Units This answer has no units #Explanation / Answer
Ans:-
Given data Pb = 10atm,Vb= 2.3*10^-3m^3
A] energy added as heat during a to b
Q = nCvT constant volume
Cv = 3/2R monoatomic gas
Q = 3/2 nRT
PV = nRT
(PV)= nRT
Q= 3/2 (PV)= 3/2(PbVb –PaVa) = 3/2(Pb-Pa)V
We need Pa =Pc and Pc+Pb connected by adiabat
PcVc^ =PbVb^
Pc = Pb(Vb/Vc)^ = 1.013*10^6Pa (1/8)^5/3
PC = 3.167*10^4 Pa =Pa
Q = 3/2 (Pb-Pa)Vb = 3/2 (1.013*10^6Pa- 3.167*10^4Pa)*(2.3*10^-3m^3)
=3.39*10^3J
B] energy leaves gas during c to a
Q = nCpT Cp = 5/2 R for monatomic
Q= 5/2nRT
= 5/2 (PV) = (PaVa – PcVc) = 5/2 Pa((Va – Vc) = 5/2Pa(-7.0Vb)
Q = 5/2 (3.167*10^4Pa) (-7*2.3*10^-3m^3 )
Q = -1274.72J
C] Eint = 0 for cyclic path so
W = Q= Qab +Qbc +Qca = 0
= |w|/|QH| = |w|/|Qab|
W = 3.39*10^3J -1274.72 = 2115.28J
= 2115.28/3.39*10^3 = 0.624
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