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Question: One mole of an ideal gas with the adiabatic exponent \(\gamma \) goes through the polytropic process...

One mole of an ideal gas with the adiabatic exponent γ\gamma goes through the polytropic process as a result of which the absolute temperature of the gas increases τ\tau fold. The polytropic constant equals nn. Find the entropy of the gas in this process.
(A) (nγ)R(n1)(γ1)lnτ\dfrac{{(n - \gamma )R}}{{(n - 1)(\gamma - 1)}}\ln \tau
(B) (nγ)R(nγ)lnτ\dfrac{{(n - \gamma )R}}{{(n - \gamma )}}\ln \tau
(C) (nγ)R(n1γ)(γ1)lnτ\dfrac{{(n - \gamma )R}}{{(n - 1\gamma )(\gamma - 1)}}\ln \tau
(D) (nγ)R(n1)γlnτ\dfrac{{(n - \gamma )R}}{{(n - 1)\gamma }}\ln \tau

Explanation

Solution

Entropy is a term in Physics that is associated with the randomness or disorderness of a particular process. In simple words, it can be explained as the heat or thermal energy that is not useful for any work, per unit temperature.
Formula used:
The formula to calculate entropy is given as
dS=dQT\int {dS = \int {\dfrac{{dQ}}{T}} }
Also, the relation between   dQ\;dQ and TT is given by,
dQ=nCdTdQ = nCdT
For a polytropic process, the formula for molar heat capacity is given as,
C=(Rγ1Rn1)C = \left( {\dfrac{R}{{\gamma - 1}} - \left. {\dfrac{R}{{n - 1}}} \right)} \right.

Complete step by step solution:
Let the initial temperature be Ti{T_i}.
It is given that the absolute temperature increases by τ\tau fold.
If the final temperature is Tf{T_f} , then we can say that,
Tf=τTi{T_f} = \tau {T_i}

This is the PVP - V curve for an increase in temperature for an adiabatic process.
From the figure, we can see that the temperature from point (1)(1) has increased to τ\tau times the initial temperature at point (2)(2).
At point (2)(2) the pressure increases and the volume decreases.
We know that the randomness increases with the decrease in volume.
Therefore we can say that the entropy increases at point (2)(2).
The formula to calculate entropy is given as
dS=dQT\int {dS = \int {\dfrac{{dQ}}{T}} }
where, dSdS is the change in entropy,
dQdQ is the heat energy released, and
TT is the temperature.
ΔS=dQT\therefore \Delta S = \int {\dfrac{{dQ}}{T}} ……….(I)(I)
Also, the relation between   dQ\;dQ and TT is given by,
dQ=nCdTdQ = nCdT
where, nn is the number of moles, and
CC is the molar heat capacity.
The value of CC can be found using specific heat capacity, or heat capacity per unit mass, c=Cmc = \dfrac{C}{m}.
From the question, we have, nn as one mole.
dQ=CdT\therefore dQ = CdT
Substituting this term in the equation (I)(I) we get,
ΔS=CdTT\Delta S = \int {\dfrac{{CdT}}{T}}
According to the requirement of the question, if we integrate the above equation from Ti{T_i} to Tf{T_f} we get, ΔS=TiTfCdTT\Delta S = \int\limits_{{T_i}}^{{T_f}} {\dfrac{{CdT}}{T}}
Upon solving the integration we get,
ΔS=ClnT]TiTf\Delta S = C\left. {\ln T} \right]_{{T_i}}^{{T_f}}
Substituting the values of the integrating limits we get,
ΔS=C(lnτTilnTi)\Delta S = C(\ln \tau {T_i} - \ln {T_i})
Using the property of logarithm, lnAlnB=lnAB\ln A - \ln B = \ln \dfrac{A}{B} we get,
ΔS=ClnτTiTi\Delta S = C\ln \dfrac{{\tau {T_i}}}{{{T_i}}}
Cancelling Ti{T_i} from numerator and denominator we get,
ΔS=Clnτ\Delta S = C\ln \tau ……….(II)(II)
For a polytropic process, the formula for molar heat capacity is given as,
C=(Rγ1Rn1)C = \left( {\dfrac{R}{{\gamma - 1}} - \left. {\dfrac{R}{{n - 1}}} \right)} \right.
where, nn is given in the question as the polytropic constant,
RR is the Rydberg’s constant, and
γ\gamma is the adiabatic exponent.
Substituting this value in the equation (II)(II) we get,
ΔS=(Rγ1Rn1)lnτ\Delta S = \left( {\dfrac{R}{{\gamma - 1}} - \left. {\dfrac{R}{{n - 1}}} \right)} \right.\ln \tau
Taking RR common,
ΔS=R(1γ11n1)lnτ\Delta S = R\left( {\dfrac{1}{{\gamma - 1}} - \left. {\dfrac{1}{{n - 1}}} \right)} \right.\ln \tau
ΔS=R[(n1)(γ1)(n1)(γ1)]lnτ\Rightarrow \Delta S = R\left[ {\dfrac{{(n - 1) - (\gamma - 1)}}{{(n - 1)(\gamma - 1)}}} \right]\ln \tau
Upon solving the numerator we get,
ΔS=R[nγ(n1)(γ1)]lnτ\Delta S = R\left[ {\dfrac{{n - \gamma }}{{(n - 1)(\gamma - 1)}}} \right]\ln \tau
ΔS=(nγ)R(n1)(γ1)lnτ\Rightarrow \Delta S = \dfrac{{(n - \gamma )R}}{{(n - 1)(\gamma - 1)}}\ln \tau

Hence the correct answer is option (A) (nγ)R(n1)(γ1)lnτ\dfrac{{(n - \gamma )R}}{{(n - 1)(\gamma - 1)}}\ln \tau .

Note: The adiabatic exponent, or the heat capacity ratio, or Laplace’s coefficient γ\gamma is equal to the ratio between the heat capacity at constant pressure, CP{C_P} and the heat capacity at constant volume, CV{C_V}. This concept is specially applied for processes involving ideal gases.