The working substance of a heat engine is 1.00 mol of a monatomic ideal gas. (1)
ID: 2233134 • Letter: T
Question
The working substance of a heat engine is 1.00 mol of a monatomic ideal gas. (1) The cycle begins at P1 = 1.00 atm and V1 = 26.8 L. (2) The gas is heated at constant volume to P2 = 2.00 atm. (3) It then expands at constant pressure until it is 53.6 L. (4) The gas is then cooled at constant volume until its pressure is again 1.00 atm. It is then compressed at constant pressure to its original state. All the steps are quasi-static and reversible.
(a) Show this cycle on a PV diagram. For each step of the cycle, find the work done by the gas, the heat absorbed by the gas, and the change in the internal energy of the gas.
W1-2 = kJ
Q1-2 = kJ
?Eint,1-2 = kJ
W2-3 = kJ
Q2-3 = kJ
?Eint,2-3 = kJ
W3-4 = kJ
Q3-4 = kJ
?Eint,3-4 = kJ
W4-1 = kJ
Q4-1= kJ
?Eint,4-1 = kJ
(b) Find the efficiency of the cycle. %
Explanation / Answer
For monoatomic gas, specific heat ratio, k = 1.67 A) On PV diagram, cycle will look like rectangle. W1-2 = 0 (constant volume process) Q1-2 = n*Cv*(T2 - T1) = n*Cv*[(P2*V2/nR) - (P1*V1/nR)] = Cv/R [P2*V2 - P1*V1] = [P2*V2 - P1*V1] / (k-1) Q1-2 = [2*10^5 * 26.8*10^-3 - 1*10^5 * 26.8*10^-3] / (1.67 - 1) = 4000 J = 4 kJ E1-2 = Q12 - W12 = 0 W23 = P2*(V3 - V2) = 2*10^5 *(53.6 - 26.8)*10^-3 = 5360 J = 5.36 kJ Q23 = n*Cp*(T3 - T2) = n*Cp*[(P3*V3/nR) - (P2*V2/nR)] = k/(k-1) *[P3*V3 - P2*V2] = 1.67/(1.67 - 1) *W23 = 13.36 kJ E23 = Q23 - W23 = 13.36 - 5.36 = 8 kJ W34 = 0 (constant volume process) Q34 = [P4*V4 - P3*V3] / (k-1) = [53.6*10^-3 *(1-2)*10^5] / (1.67-1) = -8 kJ E34 = Q34 - W34 = -8 - 0 = -8 kJ W41 = P1*(V1 - V4) = 1*10^5 *(26.8 - 53.6)*10^-3 = -2.68 kJ Q41 = k/(k-1) *W41 = 1.67/(1.67-1) *(-2.68) = -6.68 kJ E41 = Q41 - W41 = -6.68 - (-2.68) = -4 kJ b) Efficiency = (W12 + W23 + W34 + W41) / (Q12 + Q23) = (0 + 5.36 + 0 -2.68) / (4 + 13.36) = 15.4 %
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