A monatomic gas undergoes the cyclic process shown in the figure below. The temp

ID: 2071333 • Letter: A

Question

A monatomic gas undergoes the cyclic process shown in the figure below. The temperature at point C is -10 C.

(a) What is the NET work done BY the gas during one cycle? J

(b) What is the total heat input into the gas during one cycle? J

(d) What is the maximum Carnot efficiency for an engine run between the minimum and maximum temperatures for this cycle?

(e) What is the actual effiency of this engine? J

P (atm) 5.00 1.00 10.0 B V (liters 50.0

For monoatomic gas, Cv = 12.5 J/mol-K and Cp = 20.8 J/mol-K

No. of moles n = PcVc/RuTc = (1.01325*105)*(10*10-3)/(8.314*(273-10)) = 0.463

(a) Work done = area of triangle = 1/2*((50-10)*10-3)((5-1)*1.01325*105) = 8106 J

(b) For process CA, Pa/Ta = Pc/Tc

So, Ta = (5/1)*(273-10) = 1315 K

For process BC, Vb/Tb = Vc/Tc

So, Tb = (50/10)*(273-10) = 1315 K

Since Ta = Tb, change in internal energy in process AB = 0.

So, Qab = Wab = Area under line AB = 1/2*((50-10)*10-3)((5+1)*1.01325*105) = 12159 J

For BC, W = Area under BC = p(Vc-Vb) = (1.01325*105)*((10-50)*10-3) = -4053 J

For BC, U = nCv(Tc-Tb) = 0.463*12.5*((273-10)-1315) = -6088.45 J

For BC, Q = W+U = -4053 - 6088.45 = -10141.45 J

For CA: W = 0 since constant volume process.

Q = U = nCv(Ta-Tc) = 0.463*12.5*(1315-(273-10)) = 6088.45 J

So, Total heat input = 12159 + 6088.45 = 18247.45 J

(d) Carnot efficieny = 1 - T_low/T_high = 1 - (273-10)/1315 = 0.8 (or 80%)

(e) Actual efficiency = W/Heat supplied = 8106/18247.45 = 0.4442 (or 44.42%)

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