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

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.