The working substance of a heat engine is 1.00 mol of a monatomic ideal gas. (1)
ID: 3308178 • 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 = 27.4 L. (2) The gas is heated at constant volume to P2 = 2.00 atm. (3) It then expands at constant pressure until it is 54.8 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. (Do this on paper. Your instructor may ask you to turn in this 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.
(b) Find the efficiency of the cycle.
_____%
Explanation / Answer
A)
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 * 27.4*10^-3 - 1*10^5 * 27.4*10^-3] / (1.67 - 1)
= 4089.55 J
or
= 4.08 kJ
E1-2 = Q12 - W12 = 4.08955
W23 = P2*(V3 - V2)
= 2*10^5 *(54.8 - 27.4)*10^-3
= 5480 J = 5.48 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 = 13.65 kJ
E23 = Q23 - W23 = 13.65 - 5.48 = 8.17 kJ
W34 = 0 (constant volume process)
Q34 = [P4*V4 - P3*V3] / (k-1) = [54.8*10^-3 *(1-2)*10^5] / (1.67-1) = -8.179 kJ
E34 = Q34 - W34 = -8.179 - 0 = -8.179 kJ
W41 = P1*(V1 - V4) = 1*10^5 *(27.4 - 54.8)*10^-3 = -2.74 kJ
Q41 = k/(k-1) *W41 = 1.67/(1.67-1) *(-2.74) = -6.82 kJ
E41 = Q41 - W41 = -6.82 - (-2.74) = -4.08 kJ
B ) Efficiency = (W12 + W23 + W34 + W41) / (Q12 + Q23) *100
= (0 + 5.48 + 0 -2.74) / (4.08 + 13.65) *100 = 15.4 %
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