Question

In an adiabatic process, the pressure is increased by \[\dfrac{2}{3}\% .\] If \[\gamma = \dfrac{3}{2},\] then find the decrease in volume (approximately)?

Answer 1

Hint: An adiabatic process can be defined as a thermodynamic process wherein there is no exchange of heat from the system to its surroundings neither during compression nor expansion.

Formula used: The equation for adiabatic process can be written as:
\[P{V^\gamma } = K\]
Where,
\[{\text{P = }}\] Pressure of the system
\[{\text{V = }}\] Volume of the system
\[\gamma = \] Adiabatic index. It is defined as the ratio of heat capacity at constant pressure \[{C_P}\] to heat capacity at constant volume \[{C_V}.\]

Complete step by step answer:
Given:
\[{P_i} = {\text{ }}P\]
\[{P_f}{\text{ }} = {\text{ }}P{\text{ }} + {\text{ }}\dfrac{2}{3}P{\text{ }} = {\text{ }}\dfrac{5}{3}P\]
\[{V_i}{\text{ }} = {\text{ }}V\]
\[\gamma = \dfrac{3}{2}\]
To Find: \[{V_f}\]
Let us assume \[{\text{P}}\] and \[{\text{V}}\] as initial pressure and volume respectively.
For an adiabatic process,
\[P{V^\gamma } = K\]
So, \[\dfrac{{{P_i}}}{{{P_f}}} = {(\dfrac{{{V_i}}}{{{V_f}}})^\gamma }\]
Substituting the values in the above equation, we get,
\[\dfrac{P}{{(\dfrac{5}{3}P)}}{\text{ }} = {\text{ }}{(\dfrac{{{V_f}}}{V})^{\dfrac{3}{2}}}\]
\[ \Rightarrow {(\dfrac{3}{5})^{\dfrac{2}{3}}}{\text{ }} = {\text{ }}\dfrac{{{V_f}}}{V}\]
\[ \Rightarrow {V_f}{\text{ }} = {\text{ }}0.84V\]
Hence, the decrease in volume is calculated as –
\[{V_i}{\text{ }} - {\text{ }}{V_f}{\text{ }} = {\text{ }}V{\text{ }} - {\text{ }}0.84V{\text{ }} = {\text{ }}0.16V\]

Therefore, the volume is decreased by \[0.16\] times.

Note: A student can get confused with isothermal process and adiabatic process. Isothermal process is defined as the change of a system wherein the temperature remains constant. On the other hand, in adiabatic processes, no heat transfer takes place.