Figure 18-29 shows the three-part cycle for an ideal monatomic gas, the first pa

Question

Figure 18-29 shows the three-part cycle for an ideal monatomic gas, the first part of which involves an isothermal expansion from point A, where V = 38 mL and the pressure is 2.8x105 Pa, to point B, where the pressure is lower by a factor of 3.

(Note that the bottom of the graph does not necessarily correspond to zero pressure or volume.)

(a) What is the net work done during one cycle?
J
(b) What is the efficiency of this cycle?
%
(c) What is the ideal efficiency of a heat engine operating between the highest and lowest temperatures of this cycle?
%

Explanation / Answer

Note: here, the number of moles of the gas is missing, so i take n = 1 mol.

Substitute your number of moles value instead of 1 mole you will get correct answer:

At A: T = PV/nR = (2.8x105 Pa)(38x10-6 m3)/(1 mol)(8.314 J/mol.K) = 1.28 K

a) For A-B:

W = (2.8x105 Pa)(38x10-6 m3)ln(3) = 11.7 J

U = 0 (since isothrmal)

Q = W+U = 11.7 J

For B-C:

W = (2.8x105 Pa)/3 * [(38x10-6 m3) - 3*(38x10-6 m3)] = -7.1 J

U = nCpdT = 1*20.8* [1.28/3 - 1.28] = -17.75 J

Q = W+U = -7.1 J- 17.75 J = -24.85 J

For C-A:

W = 0 (since constant volume process)

U = nCvdT = 1*12.5*[1.28 - 1.28/3] = 10.7 J

Q = W+U = 10.7 J

Total work done Wtotal = 11.7 J-7.1 J = 4.6 J

b)

Efficiency = Net work/heat input

= 4.6 J /(11.7 J+10.7 J)

= 0.205 or 20.5%

c)Ideal efficiency = 1 - (1.28/3)/1.28 = 0.67 or 67 %

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