# Why change in entropy in reversible adiabatic process is zero?

In the adiabatic expansion step of carnot cycle, adiabatic change is carried out in an insulated system. In this process volume gets increased but entropy remains constant.

Now if we look at the concept (as far as I know) of entropy, it determines the number of arrangement in which gas particles can remain within the container/box. Since the volume increased, shouldn't this number of arrangement increase and a change in entropy take place? I know there is no exchange of heat but as far as I know entropy is concerned with the number of arrangement. So why entropy of an isentropic process remains constant?

The temperature also decreases due to the system doing work on the surrounding. So you have more volume but less energy throughout that volume. Entropy is not only concerned with spatial configurations.

• If you don't mind, can you provide me a source where I can learn about the whole concept of entropy upto a basic level. Most of the websites I have visited or articles I have read only explained the spatial configuration part.
– MSKB
Commented Apr 11, 2021 at 15:12

In classical statistical mechanics the entropy is related to the volume of phase space, that is the space of the positions and momentum. The volume of this space is related to the number of microstates, and it's proportional to the volume of the container, but also to the volume in momentum space. In this case the space volume increases, but the temperature (that is a measure of the mean kinetic energy) decreases, affecting the momentum space.

• Can I get an intuitive idea how does spatial arrangement and entropy relate to mean kinetic energy?
– MSKB
Commented Apr 11, 2021 at 21:28
• This is a simplified description. One container is composed by three boxes, separeted by two wall. There are 2 particles. These particles can be arranged in 9 ways, as you can see making a sketch of the system. Or thinking that it's a combinatorial problem. This is the number of way we can distribute these particles in that volume. Clearly if we increase the volume in our imaginary system, adding others boxes, the number of configurations increases and so the entropy Commented Apr 12, 2021 at 1:01
• Now, we take into account the energy. Immagine that each particle can have only two energy $E_1$ and $E_2$. How many ways there are for distributing these energies among the 2 particles? Kepting the total energy fixed. It's again combinatorial calculus. Decreasing the total energy, means decreasing the total number of way in which you can distribute this energy. So increasing or decreasing the energy, kepting other variables fixed, increases or decreases the entropy. Commented Apr 12, 2021 at 1:05
• if the mean kinetic energy is say about 100J, and every particles do nit have equal energy so there should be atleast one particle whose energy(kinetic) is 70J. In particular spatial arrangement which particle possess that 70J kinetic energy cannot be exacrly deternined but can be assumed in a number of ways. This number of arrangement of energy in a particular spatial arrangement is also a determiner of entropy. Is this notion correct? @Mark_Bell
– MSKB
Commented Apr 12, 2021 at 7:26
• In classical description you can distinguish the particles, while in quantum mechanics there is indistinguishability. Then the counting is different. Take my example of the two particles: if you cannot distinguish the particles, say for example that one particle is in box 1 and the other in box 2. Then if you exchange their position you have the same state. Commented Apr 12, 2021 at 11:17

For an ideal gas $$PV = n RT$$, the expression of entrop can be calculate:

\begin{align} dS =& \frac{dU}{T} + \frac{PdV}{T};\\ =& n C_v \frac{dT}{T} + nR\frac{dV}{V};\\ \end{align}

Carry out the integal from state 1 to state 2: \begin{align} S_2 - S_1 =& n C_v \left(\ln T_2 - \ln T_1\right) + nR \left(\ln V_2 - \ln V_1\right);\\ =& nC_v \ln \frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1}. \tag{1} \end{align}

In an adiabatic expanion, $$V_2 > V_1$$, the second term in Eq.(1) is positive; but the temperature becomes lower $$T_2 < T_1$$, the first term is negative. It is a good practice to show explicitly that these tow terms cancel each other in an adiabatic expansion process.