Question

In an adiabatic process, the pressure is increased by $\dfrac{2}{3}\%.$ If $\gamma =\dfrac{3}{2},$ then the volume decreases by nearly
$\left( a \right)\dfrac{4}{9}\%$
$\left( b \right)\dfrac{2}{3}\%$
$\left( c \right)1\%$
$\left( d \right)\dfrac{9}{4}\%$

Answer 1

We know that, in the adiabatic process, the heat remains constant, therefore the ideal gas equation for the adiabatic process is PV’ = constant. Then put the value of the ratio of specific heat $\left( \chi \right).$ Then take log on both sides and then differentiate the log equation. After that rearrange the equation, the pressure term on one side and the volume term on the other side. Then put the value of the pressure which is asked in the question in percentile form and get the value of the volume in percentile form.

Formulas Used:
The ideal gas equation for the adiabatic process is
$P{{V}^{\chi }}=\text{constant}$
Where P is pressure, V is volume and $\chi$ is the ratio of specific heat (ordinary or major)

Complete step by step answer:
We know that in an adiabatic process, the system is insulated (coated with insulating material) from the surroundings and heat absorbed or released is zero, this is a case of adiabatic where (in short) heat remains constant. Therefore, the ideal gas equation for the adiabatic equation which is given by
$P{{V}^{\chi }}=\text{constant}.....\left( i \right)$
We have been given the value of the ratio of specific heats, i.e. $\chi =\dfrac{3}{2}.$
Putting this value in equation (i), we get,
$\Rightarrow P{{V}^{\dfrac{3}{2}}}=\text{constant}$
Now, take log on both the sides, we get,
$\Rightarrow \log \left( P{{V}^{\dfrac{3}{2}}} \right)=\log \left( \text{constant} \right)$
$\Rightarrow \log P+\log \left( {{V}^{\dfrac{3}{2}}} \right)=\log \left( \text{constant} \right)$
On further solving, we get,
$\Rightarrow \log P+\dfrac{3}{2}\log V=\log \left( \text{constant} \right)$
$\Rightarrow \log P+\dfrac{3}{2}\log V=\log \text{K}$
(Since we take constant = K)
Differentiating the equation, we get,
$\dfrac{\Delta P}{P}+\dfrac{3}{2}\dfrac{\Delta V}{V}=0$
(Since differentiation of the constant is zero)
$\dfrac{\Delta V}{V}=\dfrac{-2}{3}\dfrac{\Delta P}{P}$
$\Rightarrow \dfrac{\Delta V}{V}=\dfrac{-2}{3}\times \dfrac{2}{3}$
$\Rightarrow \dfrac{\Delta V}{V}=\dfrac{-4}{9}$
Hence, the volume decreases by nearly $\dfrac{4}{9}\%.$
Therefore, the correct option is (a).

Additional information:
The ideal gas equation for the adiabatic process is
$P{{V}^{\chi }}=\text{constant}$
The term $\chi$ is known as the ratio of specific heats (ordinary or molar) at constant pressure and at a constant volume which can be expressed as
$\chi =\dfrac{{{C}_{P}}}{{{C}_{V}}}$
Where ${{C}_{P}}$ is the molar specific heat at constant pressure, ${{C}_{V}}$ is the molar specific heat at constant volume.

Note:
In this question, we got our answer in the negative form, i.e. $\dfrac{-4}{9}$ but we have taken it as $\dfrac{4}{9}$ because it is just a change. So, the minus sign won’t be in consideration. Students usually get confused between the isothermal process and the adiabatic process. The isothermal process is the process in which the temperature of the system/body is kept fixed throughout the process. The ideal gas equation for the isothermal process is given by PV = constant.
In this case, the term $\chi$ is the ratio of the specific heat that is not present. So be careful while using the ideal gas equation.