A heat engine using a diatomic ideal gas goes through the following closed cycle: ∙ Isothermal compression until the volume is halved. ∙ Isobaric expansion until the volume is restored to its initial value. ∙ Isochoric cooling until the pressure is restored to its initial value.

a)η = 0.088

b) η = 0.5

Explanation:

a) The attached figure shows the P-V diagram for the process described in the exercise. According to that figure, the work during process 1-2 is equal to:

W(1-2) = -n*R*T1*ln(vi/vf) = -n*R*T1*ln(V/(V/2)) = -n*R*T1*ln(2)

the work during process 2-3 is equal to:

W(2-3) = nR*(T2-T1)

The work done during the 3-1 process equals zero, because the volume is constant. The specific heat for the molar specific heat equals:

cp = 7*R/2, where R is gas constant.

Qin = n*cp*(T2-T1) = 7*n*R/2*(T2-T1)

the efficiency of the cycle is equal to:

η = (W(1-2) + W(2,3))/Qin = (-n*R*T1*ln2 + n*R*(T2-T1))/(7*n*R/2*(T2-T1) = (2/7)*(1-(ln2/((T2/T1)-1)))

if we write the expression between volume and temperature, we have:

T2/T1 = v1/v2

T2/T1 = v/(v/2))

T2/T1 = 2

η = (2/7)*(1-(ln2/(2-1))) = 0.088

b)

The equation for efficiency of Carnot will be equal to:

η = 1-(T1/T2) = 1 - (1/2) = 0.5

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