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Calculating entropy of a cycle consisted of isothermal, isobaric and adiabatic processes

Physics Asked by Amirhosein Rezaee on January 27, 2021

In a thermodynamic cycle, consisted of three processes: isothermal, isobaric and adiabatic, where all processes are reversible:enter image description here

Is it possible to show that change of entropy in the cycle is zero, using direct relation $oint frac{dq}{T}$ ?

One Answer

Yes! Suppose we could entropy as a function of two state variables like temperature and volume.

Eg:Consider for example an ideal gas:

$$ S = nC_v ln T + nR ln V$$

So, for a cyclic process consider a shift of state variables in the form:

$$ (T_1 , V_1) to (T_2 , V_2) to (T_3 , V_3) to (T_1,V_1)$$

And corresponding entropy change

$$Delta S = sum_{i=1}^3 S(T_{i+1},V_{i+1} ) - S(T_i , V_i)$$

With/

$$ S(T_4,V_4) = S(T_1,V_1)$$

Expand out the summation and see :-)..

Correct answer by Buraian on January 27, 2021

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