%% BEGIN BY CALCULATING THE NUMERICAAL VALUE FOR \"nR\" A closed, hydrostatic, i
ID: 3279252 • Letter: #
Question
%% BEGIN BY CALCULATING THE NUMERICAAL VALUE FOR "nR"
A closed, hydrostatic, ideal-gas system is caused to follow the following reversible cycle To prevent round-off errors in the calculations, report all answers to four (4) significant figures. (a) Calculate reversible work (w), reversible heat flow (q), and Delta U (extensive: value, sign, units in liter-atm) for each path (a, b, c, d). Summarize in a neat table the results for w, q, and Delta U for each path segment and for the entire cyclic process. (b) Calculate "Efficiency" for the cyclic process: Efficiency = (net)work done by the system/heat input to the system Write the expression as a function of T_1, T_2, T_3, and T_4, and then calculate the numerical result for this process.Explanation / Answer
nR = PV/T =1*6/300 = 1/50
assume it is monoatomic gas PV^1.67 = constant
P2/P1 = (V1/V2) ^1.67
P2 = 1 *(6/1)^1.67 = 19.93 atm
T2 = P2V2/nR = 19.93*1*50
= 996.5 K
again P3 = nRT3/V3 = 1100/(1*50) = 22 atm
P4 =22* (1/6)^1.67 = 1.104 atm
T4 = P4V4/nR
= 1.104*6*50 = 331.5 T
now Wa = delta [PV]/[1- gamma] = [19.93*1-6*1]/-0.667 =- 20.89 litre atm
Ua = 1.5*delta [PV] = [19.93*1-6*1]/0.667 = 20.89 litre atm
qa = 0
similarly Wb = 0
Ub = 1.5*[19.93-22*1] = 3.105 litre atm
qb = ub = 3.105 litre atm
now again, wc = [1.104*6 - 22*1]/[1.67-1] = 22.95 litre atm
qc = 0
uc = -22.95 litre atm
wd = 0
ud = 1.5*[1*6-1.104*6] = -0.936 litre atm
qd = -0.936 litre atm
b] efficiency = total work done*100/total heat given = [wc +wa]/qd = work done in process C and A/ heat given in process b
= [ 22.95-20.89]*100/[3.105]
= 66 %
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