Answer:
The engine operates between 86.44K and 278.84K
Explanation:
Given:
Dung the isothermal expansion portion of this engine's cycle, the volume of the gas doubles, so
V2 = 2V1 and
V3 = 2V4. (V1 to V4 represent volumes)
W,out = 960J
∆V = Increment of Volume = 5.7
n = Number of Moles of Working Substance = 1.20mol
Let Th represents hot temperature
Let Tc represents cold temperature
For a Carnot engine:
V3/V4 = V2/V1 and
Work Done in the Cycle = nR((Th)ln(V2/V1) - (Tc)ln(V3/V4)) where R = 8.31 J/mol K Gas Constant
Substitute in the values
960 = 1.2 * R [Th * ln(2V1/V1) - Tc * ln(2V4/V4)]
960 = 1.2R[Th * ln(2) - Tc * ln(2)]
960 = 1.2R ln(2) (Th - Tc) ---- Divide through by 1.2 ln(2)
960/(1.2 ln(2)) = R(Th - Tc)
1154.16 = R(Th - Tc)
Th - Tc = 1154.16/8.31
Th - Tc = 192.3593387851951
Th - Tc = 192.40K ------ (1)
For the reversible adiabatic expansion:
T2 = T1*(V1/V2)^(R/Cv).
Where V2/V1 = 5.7 ----- Given
For a monatomic ideal gas, Cv = 3/2R. Tc = T2 and Th = T1
So,
Tc = Th*(1/5.7)^(R/3/2R)
Tc = Th * (1/5.7)^⅔
Tc = Th * 0.313388769773343
Tc = Th * 0.31
Tc = 0.31Th
Substitute 0.31Th for Tc in (1)
Th - 0.31Th = 192.40
0.69Th = 192.4 ---- Divide through by 0.69
Th = 192.4/0.69
Th = 278.8405797101449
Th = 278.84K ---- Approximated
Tc = 0.31 * 278.8405797101449
Tc = 86.440579710144919
Tc = 86.44K ------ Approximated
