The second law of thermodynamics that provides the criterion for the feasibility of any process. According to the second law of thermodynamics, "for any spontaneous process the entropy of the universe increases". The second law of thermodynamics put restrictions upon the direction of heat transfer and achievable efficiencies of heat engines. Subject experts have provided a detailed explanation of each question asked in I.E. Irodov book PART TWO- THERMODYNAMICS AND MOLECULAR PHYSICS - 2.4- The Second Law of Thermodynamics- Entropy. By referring to these solutions, students can achieve higher marks in the JEE exam. Students are recommended to go through these solutions so that they can be familiar with the type of questions from The Second Law of Thermodynamics.
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?
Solution:
The efficiency is given by
η = (T_{1}-T_{2})/T_{1}, (T_{1} > T_{2})
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.
Solution:
For H_{2}, γ = 7/5
p_{1}V_{1} = p_{2}V_{2} , P_{3}V_{3} = p_{4}V_{4}
p_{2}V_{2}^{γ} = p_{3}V_{3}^{ γ}_{, } P_{1}V_{1}^{ γ} = p_{4}V_{4}^{ γ}
Define n by V_{3} = nV_{2}
Then p_{3} = p_{2} n^{- γ}, so
p_{4}V_{4 }= p_{3}V_{3} = p_{2}V_{2} n^{ 1-γ} = p_{1}V_{1} n^{ 1-γ}
p_{4}V_{4}^{ γ}_{= p1V1γ}
so, V_{4}^{1-γ} = V_{1}^{1-γ} n^{ 1-γ}
or V_{4} = nV_{1}
Also,
(b) Define n by p_{3} = p_{2}/n
p_{2}V_{2}^{γ} = (p_{2}/n)V_{3}^{γ} or V_{3} = n^{1/γ} V_{2}
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.
Solution:
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 T_{1}, T_{2}, T_{3}. Find the efficiency of such a cycle, if in each isothermal expansion the gas volume increases in the same proportion.
Solution:
Given: V_{2} = nV_{1}, V_{4} = nV_{3}
Q_{1} = 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.
Solution:
on the other hand,
6p_{1}V_{1}^{γ} = p_{2}V_{2}^{γ}, p_{3}V_{2}^{γ} = p_{4}V_{1}^{γ}
Also, V_{2} = nV_{1}
Thus, p_{1} = p_{2}n^{γ} , p_{4} = p_{3}n^{γ}
η = 1 – n^{1-γ } with γ = 7/5 for N_{2}.
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.
Solution:
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.
Solution:
Since the absolute temperature of the gas rises n times both in the isochoric heating and in the isobaric expansion.
p_{1}np_{2} and V_{2} = nV_{1}
Heat taken is Q_{1} = Q_{11} + Q_{12}
Q_{11} = c_{p}(n-1)T_{1} and Q_{12 }= c_{v}(1-1/n)T_{1}
Heat rejected is Q' _{2} = Q'_{ 21} + Q’_{22} where
Q^{’}_{21} = c_{v}(n-1)T_{1} and Q^{’}_{22 }= c_{p}(1-1/n)T_{1}
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.
Solution:
(a) p_{2} = np_{1}, p_{1}V_{1} = p_{0}V_{0},
np_{1}V_{1}^{γ} = p_{0}V_{0}^{ γ}
Q’_{2} = RT_{0} ln(V_{o}/V_{1}), Q_{1} = C_{v}T_{o}(n-1)
But nV_{1}^{γ-1} = V_{0}^{ γ-1} or V_{1} = V_{0}n^{-1/(γ-1)}
Q’_{2} = RT_{0} 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) V_{2} = nV_{1}, p_{1}V_{1} = p_{0}V_{0},
And p_{1} (nV_{1}) ^{γ} = p_{0}V_{0}^{ γ}
i.e. n^{γ} V_{1}^{ γ-1} = V_{0}^{ γ-1} or V_{1} = n^{- γ/( γ-1} V_{0}
Also, Q_{1} = C_{p}T_{0}(n-1), Q’_{2} = RT_{0} ln(V_{o}/V_{1})
Or Q’_{2} = RT_{0} ln(n^{ γ/(γ-1)}) = R^{γ/(γ-1) To ln n = Cp To 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:
Solution: Here the isothermal process proceeds at the maximum temperature instead of at the minimum temperature of the cycle.
(a) p_{2} = p_{1}/n, p_{1}V_{1} = p_{0}V_{0}
p_{2}V_{1}^{γ} = p_{0}V_{0}^{ γ} or p_{1}V_{1}^{γ} = np_{0}V_{0}^{ γ}
i.e. V_{1}^{γ-1 } = nV_{0}^{γ-1} or V_{1} = V_{0}n^{1/(γ-1)}
Q’_{2} = C_{v}T_{0} (1-1/n), Q_{1} = RT_{0} ln(V_{1}/V_{0}) = [RT_{o}/(^{γ-1)] ln n = CvTo ln n.}
Thus, η = 1 – Q’_{2}/Q_{1} = 1 – (n-1)/(n ln n)
(b) p_{2} = V_{1}/n, p_{1}V_{1} = p_{0}V_{0}
P_{0}V_{o}^{γ} = p_{1}V_{2}^{ γ} = p_{1} n^{-γ} V_{1}^{γ} = V_{0}^{γ-1} n^{-γ} V_{1}^{γ-1 } or V_{1} =^{n (γ /(γ-1) V0}
Q’_{2} = C_{p}T_{0} (1-1/n), Q_{1} = RT_{0} ln(V_{1}/V_{0}) = [R γ/(^{γ-1)] To ln n = CpTo ln n.}
Thus, η = 1 – Q’_{2}/Q_{1} = 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.
Solution:
The section from (p_{1}, V_{1}, T_{0}) to (p_{2}, V_{2}, T_{0}/n) is a polytrophic process of index a. We shall assume that the corresponding specific heat C is + ve.
Here dQ = CdT = C_{v}dT + pdV
Now, pV^{α} = constant or TV^{α-1} = constant
So, pdV = (RT/V)dV = -(R/(α-1))dT
Then
We have, p_{1}V_{1} = RT_{0} = p_{2}V_{2} = RT_{o}/n = p_{1}V_{1}/n
p_{0}V_{0} = p_{1}V_{1} = n p_{2}V_{2}, p_{0}V_{0}^{γ}= p_{2}V_{2}^{ γ},
p_{1}V_{1}^{α} = p_{2}V_{2}^{ α} or V_{0}^{γ-1 }= (1/n)V_{2}^{ γ-1 }or V_{2}^{= Vo (n)1/(γ-1)}
V_{1}^{α-1 }= (1/n)V_{2}^{ α-1 } or V_{1} = n^{(-1/ α-1)}, V_{2} = (n)^{1/(γ-1) – 1/( α-1)} V_{0}
Now, Q’_{2} = CT_{0}(1 – 1/n), Q_{1} = RT_{o} ln(V_{1}/V_{0}) = RT_{o}(1/(^{γ-1) – 1/(α-1)ln n = CT0 ln n}
Thus, η = 1 – Q’_{2}/Q_{1} = 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.
Solution:
(a) Here
Along the adiabatic line, T_{o}V_{o}^{γ-1} = T_{1}(nV_{o}) ^{γ-1} or T_{o } = T_{1}n ^{γ-1}
So, Q_{1} = C_{v}(T_{1}/n)(n^{γ }-1)
Thus, η = 1 – [γ(n-1)]/[ (n^{γ -1})]
(b) Here
Along the adiabatic line, TV ^{γ-1} = constant
T_{o}V_{o}^{γ-1} = T_{1}(V_{o}/n) ^{γ-1} or T_{1 } = T_{0}n ^{γ-1}
Thus, η = 1 – [n^{γ} -1]/[^{(n-1)γnγ -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.
Solution:
(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.
Solution:
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.
Solution:
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.
Solution:
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}.
Solution:
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}/Q_{1}
η = 1 – Q’_{2}/Q_{1} < 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.
Solution:
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.
Solution:
(a) In an isochoric process the entropy change will be
For carbon dioxide, γ = 1.30
∆S = 19.2 Joule/^{o}K-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/^{o}K-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?
Solution:
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.
Solution:
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/^{o}K.
(assuming that the final volume is n times the initial volume)