Question

The entropy change in an adiabatic process is:
A) Zero
B) Always positive
C) Always negative
D) Sometimes positive and sometimes negative

Answer 1

Hint: In the adiabatic process, no heat enters or leaves the system. Therefore, the change in heat equals to zero. The change in entropy is the change in heat ($\text{dq}$) of the system per unit rise in temperature. Therefore, for the adiabatic process, the entropy change $$\mathrm{\Delta }\text{ S}$$ will always remain constant.

Complete step by step answer:
Entropy is the measure of disorder or randomness of the system. There are several other definitions of entropy. For a system, the entropy S is represented as the change in the heat ‘q’ of the system per unit of absolute temperature (T). Mathematically, the change in entropy can be depicted as follows:
$\text{Change in Entropy (}\mathrm{\Delta }\text{S)= }{\displaystyle \frac{\text{Change in heat of the system}}{\text{Temperature in Kelvin}}}\text{=}{\displaystyle \frac{\text{dq}}{\text{T}}}$
Or $$\mathrm{\Delta }\text{ S = }\int {\displaystyle \frac{\text{d}{\text{q}}_{\text{rev}}}{\text{T}}}$$
It is the ratio of change in the heat entering or leaving the system to the absolute temperature of the system.
We are interested to find out the entropy change associated with the adiabatic process.
According to thermodynamics, a process is said to be adiabatic if no heat enters or leaves the system during any stage of the process. As no heat is allowed to transfer between the surrounding and system, the heat remains constant.
Therefore, the heat change in between the system and surrounding is equalled to zero. That is,
$\text{Change in heat (q) = dq = 0}$
We know that the entropy of the system depends on the heat of the system. The entropy change for the change in the heat of the system at a constant temperature for the reversible reaction is,
$\mathrm{\Delta }\text{ S = }{\displaystyle \frac{\text{d}{\text{q}}_{\text{rev}}}{\text{T}}}$
For the adiabatic process, the entropy would be,
$\mathrm{\Delta }\text{ S = constant }\because \text{dq}=0$
Thus The change in entropy between the ${\text{S}}_{\text{1}}$ and ${\text{S}}_2$ would be equal to zero.
$\mathrm{\Delta }\text{ S = }{\text{S}}_{\text{2}}\text{- }{\text{S}}_{\text{1}}\text{ = 0}$
This is applicable for the only reversible adiabatic process. Since in the reversible processes the system covers the path and returns back to its original state, thus the net change in the heat would be equal to zero. In a reversible process, like the Carnot engine, the system may change entropy but the overall change is zero. The system does not affect the entropy of the surrounding since no heat transfer.
For the irreversible process (for example, diffusion of solute from high concentrated region to lower concentrated region), the system does not return to the original state. Therefore, the net change in the entropy is always non zero or they increase the entropy of the system.
However, in the reversible process, the changes neither change the entropy of the system nor the entropy of the surrounding leads to the zero change in entropy.
Therefore, the change in the entropy for an adiabatic process equals to zero.

Hence, (A) is the correct option.

Note: 1) For the isothermal process $\text{( }\mathrm{\Delta }\text{ T=constant)}$ , the entropy is given as,
$\mathrm{\Delta }{\text{S}}_{\text{T}}\text{=R ln}\left({\displaystyle \frac{{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}}}\right)\text{= }-\text{R ln}\left({\displaystyle \frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}}\right)$
2) For the isochoric process$\text{( }\mathrm{\Delta }\text{ P=constant)}$, the entropy is given as,
$\mathrm{\Delta }{\text{S}}_{\text{P}}\text{=}{\text{C}}_{\text{p}}\text{ ln}\left({\displaystyle \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}}\right)$
Where, ${\text{C}}_{\text{p}}$ is the specific heat of the system at constant pressure.