Question

For the process to occur under adiabatic conditions, the correct condition is:
(a) $\Delta T = 0$
(b) $\Delta p = 0$
(c) q = 0
(d) W = 0

Answer 1

Hint: In thermodynamics, an adiabatic process occurs without transferring heat or mass between a thermodynamic system and its surroundings. Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work.

Complete step by step answer:
1: We know adiabatic processes occur without the transfer of heat or mass. One example for adiabatic process would be the release of air from a pneumatic tyre. Adiabatic efficiency is also applied to devices such as nozzles, compressors and turbines. This application clearly states that there can be change in pressure (in nozzles, compressors, etc.)
2: For adiabatic process, relation between pressure and volume is given as:
\[
  P{V^\gamma } = c \\
  {P_1}{V_1}^\gamma = {P_2}{V_2}^\gamma \\
 \] where, P is pressure, V is volume, ɤ is adiabatic index, c is a constant
${(\dfrac{{{P_1}}}{{{P_2}}})^\gamma } = {(\dfrac{{{V_1}}}{{{V_2}}})^\gamma }$
If ${P_1} = {P_2}$ , then ${V_1}^\gamma = V_2^\gamma $
If ${V_1} = {V_2}$
 then no work will be done, so pressure always remains constant. So, $\Delta p = 0$is a wrong statement.
3: Also, energy is transformed as work in an adiabatic process. So, W=0 is also a wrong statement.
4: The temperature can be varied in one process, without transfer of heat, as there is exchange of energy. So, $\Delta T = 0$ is also a wrong statement.
5: Thus option (c) is the correct one as there is no transfer of heat. (q=0) Another example would be an icebox where no heat comes out or goes in and the ice doesn’t melt.
Adiabatic system must be perfectly insulated from the surroundings.

Thus, the correct option is (C).

Note:
The other conditions are applicable in different thermodynamic processes.
$\Delta T = 0$ is an isothermal process.
 $\Delta p = 0$ is an isobaric process.
W=0 will occur when isovolumetric or isometric processes take place at constant volume. (aka isochoric process) Then,$\Delta U = \Delta Q$. All the heat added to the system goes into increasing its internal energy.