Question

In an adiabatic process, the pressure is increased by $\dfrac{2}{3}$ %. If $\gamma = \dfrac{3}{2}$, then nearly by what percentage the volume decreases?
${\text{A}}{\text{.}}$ $\dfrac{4}{9}$ %
${\text{B}}{\text{.}}$ $\dfrac{2}{3}$ %
${\text{C}}{\text{.}}$ 1 %
${\text{D}}{\text{.}}$ $\dfrac{9}{4}$ %

Answer 1

Hint: Here, we will proceed by taking the natural logarithm of the general equation for any adiabatic process. Through this, we will represent the percentage change in pressure with the percentage change in volume.

Formulas Used- ${\text{P}}{{\text{V}}^\gamma }$ = c, $\ln \left( {ab} \right) = \ln a + \ln b$, $\ln \left( {{a^b}} \right) = b\ln a$ and $\ln x + a\ln y = b \Rightarrow \dfrac{{\Delta x}}{x} + a\dfrac{{\Delta y}}{y} = 0$.

Step By Step Answer:
Given, Percentage increase in the pressure = $\dfrac{2}{3}$ %

Adiabatic index $\gamma = \dfrac{3}{2}$

As we know that in any adiabatic process, ${\text{P}}{{\text{V}}^\gamma }$ = c $\to (1)$ where P is the pressure in the adiabatic process, V is the volume in the adiabatic process, $\gamma$ is the adiabatic index and c is any constant

By taking natural logarithm on both sides of equation (1), we get

$\Rightarrow \ln \left( {{\text{P}}{{\text{V}}^\gamma }} \right) = \ln \left( {\text{c}} \right)$
Using the formula $\ln \left( {ab} \right) = \ln a + \ln b$ in the above equation, we get
$\Rightarrow \ln \left( {\text{P}} \right) + \ln \left( {{{\text{V}}^\gamma }} \right) = \ln \left( {\text{c}} \right)$

By using the formula $\ln \left( {{a^b}} \right) = b\ln a$ in the above equation, we get
$\Rightarrow \ln \left( {\text{P}} \right) + \gamma \ln \left( {\text{V}} \right) = \ln \left( {\text{c}} \right)$

By putting $\gamma = \dfrac{3}{2}$ in the above equation, we get

$\Rightarrow \ln \left( {\text{P}} \right) + \dfrac{3}{2}\ln \left( {\text{V}} \right) = \ln \left( {\text{c}} \right){\text{ }} \to {\text{(2)}}$

Also we know that any equation in x and y variables given by $\ln x + a\ln y = b$ where a and b are constants can be written as

$\Rightarrow \dfrac{{\Delta x}}{x} + a\dfrac{{\Delta y}}{y} = 0$

Using the above concept, equation (1) can be written in terms of change in pressure and change in volume (these two are variables) can be written as

$\Rightarrow \dfrac{{\Delta {\text{P}}}}{{\text{P}}} + \dfrac{3}{2}\dfrac{{\Delta {\text{V}}}}{{\text{V}}} = 0{\text{ }} \to {\text{(2)}}$

As, Percentage increase in pressure = $\dfrac{{{\text{Final Pressure}} - {\text{Initial Pressure}}}}{{{\text{Initial Pressure}}}} \times 100 = \dfrac{{\Delta {\text{P}}}}{{\text{P}}} \times 100$ % where $\Delta {\text{P}}$ denotes the change in pressure

$\Rightarrow \dfrac{2}{3} = \dfrac{{\Delta {\text{P}}}}{{\text{P}}} \times 100 \\\ \Rightarrow \dfrac{{\Delta {\text{P}}}}{{\text{P}}} = \dfrac{2}{{3 \times 100}} \\\ \Rightarrow \dfrac{{\Delta {\text{P}}}}{{\text{P}}} = \dfrac{2}{{300}} \\\$

By substituting $\dfrac{{\Delta {\text{P}}}}{{\text{P}}} = \dfrac{2}{{300}}$ in equation (2), we get

$$\Rightarrow \dfrac{2}{{300}} + \dfrac{3}{2}\dfrac{{\Delta {\text{V}}}}{{\text{V}}} = 0 \\\ \Rightarrow \dfrac{3}{2}\dfrac{{\Delta {\text{V}}}}{{\text{V}}} = - \dfrac{2}{{300}} \\\ \Rightarrow \dfrac{{\Delta {\text{V}}}}{{\text{V}}} = - \dfrac{{2 \times 2}}{{300 \times 3}} \\\ \Rightarrow \dfrac{{\Delta {\text{V}}}}{{\text{V}}} = - \dfrac{4}{{900}} \\\$$

Also, Percentage change in volume = $\dfrac{{{\text{Final Volume}} - {\text{Initial Volume}}}}{{{\text{Initial Volume}}}} \times 100 = \dfrac{{\Delta {\text{V}}}}{{\text{V}}} \times 100$ % where $\Delta {\text{V}}$ denotes the change in volume

Percentage change in volume = $- \dfrac{4}{{900}} \times 100 = - \dfrac{4}{9}$ %

The negative sign of the percentage change in volume means that the volume is decreased.
Therefore, the percentage decrease in volume is $\dfrac{4}{9}$ %

Hence, option A is correct.

Note- In this particular problem, P and V are considered as two state variables which are basically varied as the state is changed. Also, the percentage change in any quantity is defined as the ratio of the change in the value of that quantity to the original value of that quantity multiplied by 100. This percentage change can be percentage increase or percentage decrease.