Question

The working substance of a certain Carnot engine is 1.20 mol of an ideal monatomic gas. During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles, while during the adiabatic expansion the volume increases by a factor of 5.7. The work output of the engine is 960 J in each cycle.Compute the temperatures of the two reservoirs between which this engine operates.

Answer 1

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