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

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 $$\oint \frac{dq}{T}$$ ?

• 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. Dec 22, 2020 at 18:20
• Nevertheless, shouldn't these two methods be consistent to each other? Dec 22, 2020 at 18:28
• It depends. What is the exact statement of the problem? It matters how much information (input data) you are given. Dec 22, 2020 at 20:50

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 :-)..