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

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.