I.E. Irodov Solutions on The Second Law of Thermodynamics- Entropy

1. In which case will the efficiency of a Carnot cycle be higher: when the hot body temperature is increased by ΔT, or when the cold body temperature is decreased by the same magnitude?

The efficiency is given by

η = (T₁-T₂)/T₁, (T₁ > T₂)

Now in the two cases the efficiencies are

Thus, η_k < η_l

2. Hydrogen is used in a Carnot cycle as a working substance. Find the efficiency of the cycle, if as a result of an adiabatic expansion (a) the gas volume increases n = 2.0 times; (b) the pressure decreases n = 2.0 times.

For H₂, γ = 7/5

p₁V₁ = p₂V₂ , P₃V₃ = p₄V₄

p₂V₂ ^γ = p₃V₃ ^γ_, P₁V₁ ^γ = p₄V₄ ^γ

Define n by V₃ = nV₂

Then p₃ = p₂ n^{– γ}, so

p₄V₄ = p₃V₃ = p₂V₂ n ^1-γ = p₁V₁ n ^1-γ

p₄V₄ ^γ = p₁V₁^γ

so, V₄^1-γ = V₁^1-γ n ^1-γ

or V₄ = nV₁

Also,

(b) Define n by p₃ = p₂/n

p₂V₂^γ = (p₂/n)V₃^γ or V₃ = n^1/γ V₂

If n approaches to n^1/γ in the previous case

Η = 1-n^1/γ-1 = 1 – n^-2/7 = 0.18(approx.)

3. A heat engine employing a Carnot cycle with an efficiency of η = 10% is used as a refrigerating machine, the thermal reservoirs being the same. Find its refrigerating efficiency ε.

Used as a refrigerator, the refrigerating efficiency of a heat engine is given by

where η is the efficiency of the heat engine.

4. An ideal gas goes through a cycle consisting of alternate isothermal and adiabatic curves (below Fig.). The isothermal processes proceed at the temperatures T₁, T₂, T₃. Find the efficiency of such a cycle, if in each isothermal expansion the gas volume increases in the same proportion.

Given: V₂ = nV₁, V₄ = nV₃

Q₁ = Heat taken at the upper temperature

5. Find the efficiency of a cycle consisting of two isochoric and two adiabatic lines, if the volume of the ideal gas changes n = 10 times within the cycle. The working substance is nitrogen.

on the other hand,

6p₁V₁^γ = p₂V₂^γ, p₃V₂^γ = p₄V₁^γ

Also, V₂ = nV₁

Thus, p₁ = p₂n^γ , p₄ = p₃n^γ

η = 1 – n^1-γ with γ = 7/5 for N₂.

6. Find the efficiency of a cycle consisting of two isobaric and two adiabatic lines, if the pressure changes n times within the cycle. The working substance is an ideal gas whose adiabatic exponent is equal to γ.

7. An ideal gas whose adiabatic exponent equals γ goes through a cycle consisting of two isochoric and two isobaric lines. Find the efficiency of such a cycle, if the absolute temperature of the gas rises n times both in the isochoric heating and in the isobaric expansion.

Since the absolute temperature of the gas rises n times both in the isochoric heating and in the isobaric expansion.

p₁np₂ and V₂ = nV₁

Heat taken is Q₁ = Q₁₁ + Q₁₂

Q₁₁ = c_p(n-1)T₁ and Q₁₂ = c_v(1-1/n)T₁

Heat rejected is Q’ ₂ = Q’ ₂₁ + Q’₂₂ where

Q^’₂₁ = c_v(n-1)T₁ and Q^’₂₂ = c_p(1-1/n)T₁

Thus,

8. An ideal gas goes through a cycle consisting of (a) isochoric, adiabatic, and isothermal lines; (b) isobaric, adiabatic, and isothermal lines, with the isothermal process proceeding at the minimum temperature of the whole cycle. Find the efficiency of each cycle if the absolute temperature varies n-fold within the cycle.

(a) p₂ = np₁, p₁V₁ = p₀V₀,

np₁V₁ ^γ = p₀V₀ ^γ

Q’₂ = RT₀ ln(V_o/V₁), Q₁ = C_vT_o(n-1)

But nV₁ ^γ-1 = V₀ ^γ-1 or V₁ = V₀n^-1/(γ-1)

Q’₂ = RT₀ ln(n^1/(γ-1)) = [RT_o/( γ-1)] ln n

Thus, η = 1 – ln n/(n-1) on using C_v = R/( γ-1)

(b) V₂ = nV₁, p₁V₁ = p₀V₀,

And p₁ (nV₁) ^γ = p₀V₀ ^γ

i.e. n^γ V₁ ^γ-1 = V₀ ^γ-1 or V₁ = n^{– γ/( γ-1} V₀

Also, Q₁ = C_pT₀(n-1), Q’₂ = RT₀ ln(V_o/V₁)

Or Q’₂ = RT₀ ln(n ^γ/(γ-1)) = R γ/( γ-1) T_o ln n = C_p T_o ln n

Thus, η = 1 – (ln n)/(n-1)

9. The conditions are the same as in the foregoing problem with the exception that the isothermal process proceeds at the maximum temperature of the whole cycle.

Solution: Here the isothermal process proceeds at the maximum temperature instead of at the minimum temperature of the cycle.

(a) p₂ = p₁/n, p₁V₁ = p₀V₀

p₂V₁ ^γ = p₀V₀ ^γ or p₁V₁ ^γ = np₀V₀ ^γ

i.e. V₁ ^γ-1 = nV₀^γ-1 or V₁ = V₀n^1/(γ-1)

Q’₂ = C_vT₀ (1-1/n), Q₁ = RT₀ ln(V₁/V₀) = [RT_o/( γ-1)] ln n = C_vT_o ln n.

Thus, η = 1 – Q’₂/Q₁ = 1 – (n-1)/(n ln n)

(b) p₂ = V₁/n, p₁V₁ = p₀V₀

P₀V_o ^γ = p₁V₂ ^γ = p₁ n^-γ V₁ ^γ = V₀^γ-1 n^-γ V₁ ^γ-1 or V₁ = n ^{(γ /(γ-1)} V₀

Q’₂ = C_pT₀ (1-1/n), Q₁ = RT₀ ln(V₁/V₀) = [R γ/( γ-1)] T_o ln n = C_pT_o ln n.

Thus, η = 1 – Q’₂/Q₁ = 1 – (n-1)/(n ln n)

10. An ideal gas goes through a cycle consisting of isothermal, polytrophic, and adiabatic lines, with the isothermal process proceeding at the maximum temperature of the whole cycle. Find the efficiency of such a cycle if the absolute temperature varies n-fold within the cycle.

The section from (p₁, V₁, T₀) to (p₂, V₂, T₀/n) is a polytrophic process of index a. We shall assume that the corresponding specific heat C is + ve.

Here dQ = CdT = C_vdT + pdV

Now, pV^α = constant or TV^α-1 = constant

So, pdV = (RT/V)dV = -(R/(α-1))dT

Then

We have, p₁V₁ = RT₀ = p₂V₂ = RT_o/n = p₁V₁/n

p₀V₀ = p₁V₁ = n p₂V₂, p₀V₀ ^γ= p₂V₂ ^γ,

p₁V₁ ^α = p₂V₂ ^α or V₀ ^γ-1 = (1/n)V₂ ^γ-1 or V₂ = V_o (n)^1/(γ-1)

V₁^α-1 = (1/n)V₂ ^α-1 or V₁ = n^{(-1/ α-1)}, V₂ = (n)^{1/(γ-1) – 1/( α-1)} V₀

Now, Q’₂ = CT₀(1 – 1/n), Q₁ = RT_o ln(V₁/V₀) = RT_o(1/( γ-1) – 1/( α-1)ln n = CT₀ ln n

Thus, η = 1 – Q’₂/Q₁ = 1 – (n-1)/(n ln n)

11. An ideal gas with the adiabatic exponent γ goes through a direct (clockwise) cycle consisting of adiabatic, isobaric, and isochoric lines. Find the efficiency of the cycle if in the adiabatic process the volume of the ideal gas (a) increases n-fold; (b) decreases n-fold.

(a) Here

Along the adiabatic line, T_oV_o ^γ-1 = T₁(nV_o) ^γ-1 or T_o = T₁n ^γ-1

So, Q₁ = C_v(T₁/n)(n^γ -1)

Thus, η = 1 – [γ(n-1)]/[ (n^{γ -1})]

(b) Here

Along the adiabatic line, TV ^γ-1 = constant

T_oV_o ^γ-1 = T₁(V_o/n) ^γ-1 or T₁ = T₀n ^γ-1

Thus, η = 1 – [n^γ -1]/[ (n-1)γn^{γ -1}]

12. Calculate the efficiency of a cycle consisting of isothermal, isobaric, and isochoric lines, if in the isothermal process the volume of the ideal gas with the adiabatic exponent γ.

(a) increases n-fold (b) decreases n-fold.

(b) Here

Thus, η = 1 – [(n -1)+ (γ-1)ln n]/[ (n-1)γ]

13. Find the efficiency of a cycle consisting of two isochoric and two isothermal lines if the volume varies v-fold and the absolute temperature τ-cycle within the cycle. The working substance is an ideal gas with the adiabatic exponent γ.

14. Find the efficiency of a cycle consisting of two isobaric and two isothermal lines if the pressure varies n-fold and the absolute temperature τ-fold within the cycle. The working substance is an ideal gas with the adiabatic exponent γ.

15. An ideal gas with the adiabatic exponent γ goes through a cycle (Fig. below) within which the absolute temperature varies τ-fold. Find the efficiency of this cycle.

Due to linearity of the section

BC whose equation is

p/p_o = vV/V_o

we have

16. Making use of the Clausius inequality, demonstrate that all cycles having the same maximum temperature T_max and the same minimum temperature T_min are less efficient compared to the Carnot cycle with the same T_max and T_min.

Claussius inequality can be written in the form:

In this inequality, T_max > T > T_min and we can write

Thus, T_min/T_max < Q’₂/Q₁

η = 1 – Q’₂/Q₁ < 1 – T_min/T_max = η_carnot

17. Making use of the Carnot theorem, show that in the case of a physically uniform substance whose state is defined by the parameters T and V,

Where U (T, V) is the internal energy of the substance.

Instruction: Consider the infinitesimal Carnot cycle in the variables p, V.

We consider an infinitesimal carnot cycle with isothermal process at temperatures T + dT and T. Let δA be the work done in the cycle and δQ, be the heat received at the higher temperature. Then by Carnot’s theorem

18. Find the entropy increment of one mole of carbon dioxide when its absolute temperature increases n = 2.0 times if the process of heating is (a) isochoric; (b) isobaric. The gas is to be regarded as ideal.

(a) In an isochoric process the entropy change will be

For carbon dioxide, γ = 1.30

∆S = 19.2 Joule/^oK-mole.

(b) For an isobaric process,

∆S = C_p ln (T_f/T_i ) = C_p ln n = [γR ln n]/[ γ-1] = 25 Joule/^oK-mole.

19. The entropy of v = 4.0 moles of an ideal gas increases by ΔS = 23 J/K due to the isothermal expansion. How many times should the volume v = 4.0 moles of the gas be increased?

In an isothermal expansion

∆S = vR ln (V_f/V_i )

V_f/V_i = e^∆S/VR = 2.0 times

20. Two moles of an ideal gas are cooled isochorically and then expanded isobarically to lower the gas temperature back to the initial value. Find the entropy increment of the gas if in this process the gas pressure changed n = 3.3 times.

The entropy change depends on the final & initial states only, so we can calculate it directly along the isotherm, it is ∆S = 2R ln n = 20 J/^oK.

(assuming that the final volume is n times the initial volume)