In a thermodynamic cycle, consisted of three processes: isothermal, isobaric and adiabatic, where all processes are reversible:
Is it possible to show that change of entropy in the cycle is zero, using direct relation $\oint \frac{dq}{T}$ ?
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1$\begingroup$ You are thinking about it backwards. Entropy is a function of state, and since, in a cyclic process, the working fluid returns to its original state at the end of each cycle, the change in entropy for the working fluid over a cycle must be zero. $\endgroup$– Chet MillerDec 22, 2020 at 18:20
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1$\begingroup$ Nevertheless, shouldn't these two methods be consistent to each other? $\endgroup$– Amirhossein RezaeiDec 22, 2020 at 18:28
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$\begingroup$ It depends. What is the exact statement of the problem? It matters how much information (input data) you are given. $\endgroup$– Chet MillerDec 22, 2020 at 20:50
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1 Answer
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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 :-)..
answered Dec 22, 2020 at 18:48