(a) For 3.05 moles of a monatomic ideal gas taken through the cycle in the figur
ID: 3278016 • Letter: #
Question
(a) For 3.05 moles of a monatomic ideal gas taken through the cycle in the figure, where V_1 = 5.00V_0 what is W/p_0 V_0 as the gas goes from state a to state c along path abc? What is delta E_int/p_0 V_0 in going (b) from b to c and (c) through one full cycle? What is delta S in going (d) from b to c and (e) through one full cycle? (a) Number 2 Units This answer has no units (b) Number 15 Units This answer has no units (c) Number 0 Units J (d) Number 8.314 Units J/kg middot K (e) Number 0 Units J/KExplanation / Answer
a. given, number of moles, n = 3.05
monoatomic ideal gas
V1 = 5Vo
SO, From a to b
Work done, W = po(V1 - Vo) = po(4Vo) = 4PoVo
From b to C
Work done, W = 0 [ isochoric process]
so, W/povo = 4
b. from b to c ( isochoric process )
W = 0
dEint = nCv(Tc - Tb)
now, PbVb = nRTb
Tb = 5*poVo/nR
and similiaarly, Tc = 10poVo/nR
and Cv for monoatomic gas = 3R/2
so, dEint = 3.05*3R(10 - 5)poVo/2*3.05*R = 15poVo/2
dEint/povo = 7.5
c. dEint for full cycle = 0 ( closed loop)
so, dEint/povo = 0
d. entropy change in isochoric process is given by
dS = nCvln(T2/T1) = 3.05*3Rln(2)/2 = 26.352 J/K
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