Question:

A monoatomic gas having $ \gamma = \frac{5}{3} $ is stored in a thermally insulated container and the gas is suddenly compressed to $ \left( \frac{1}{8} \right)^{\text{th}} $ of its initial volume. The ratio of final pressure and initial pressure is:

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In adiabatic processes, the relationship between pressure and volume follows the equation $ P_1 V_1^\gamma = P_2 V_2^\gamma $. When the volume changes, you can find the new pressure by applying this equation with the given values for $ V_1 $, $ V_2 $, and $ \gamma $.

Updated On: Jan 31, 2026

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The Correct Option is B

Approach Solution - 1

To solve this problem, we need to make use of the adiabatic process relation for an ideal monoatomic gas. In an adiabatic process, the relationship between pressure and volume is given by:

$P_1 V_1^\gamma = P_2 V_2^\gamma$

Where:

$P_1$ and $V_1$ are the initial pressure and volume, respectively.
$P_2$ and $V_2$ are the final pressure and volume, respectively.
$\gamma$ (gamma) is the adiabatic index, which is given as $\frac{5}{3}$ for a monoatomic gas.

From the problem, the gas is compressed to $\left(\frac{1}{8}\right)^{\text{th}}$ of its initial volume. Thus, we have:

$V_2 = \frac{V_1}{8}$

Substituting the values into the adiabatic equation, we get:

$P_1 V_1^{\frac{5}{3}} = P_2 \left(\frac{V_1}{8}\right)^{\frac{5}{3}}$

We can simplify this equation as follows:

$P_1 V_1^{\frac{5}{3}} = P_2 \cdot \frac{V_1^{\frac{5}{3}}}{8^{\frac{5}{3}}}$

$P_1 = P_2 \cdot \frac{1}{8^{\frac{5}{3}}}$

Rearranging for $\frac{P_2}{P_1}$, we get:

$\frac{P_2}{P_1} = 8^{\frac{5}{3}}$

Calculating the value of $8^{\frac{5}{3}}$:

$8 = 2^3$, so $8^{\frac{5}{3}} = (2^3)^{\frac{5}{3}} = 2^5 = 32$

Therefore, the ratio of final pressure to initial pressure is 32.

Hence, the correct answer is 32.

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Approach Solution -2

For a thermally insulated (adiabatic) process, the relation between pressure and volume is given by: \[ P_1 V_1^\gamma = P_2 V_2^\gamma \] Where: - $ P_1 $ and $ V_1 $ are the initial pressure and volume, - $ P_2 $ and $ V_2 $ are the final pressure and volume, - $ \gamma = \frac{5}{3} $ is the adiabatic index (ratio of specific heats). Given that the gas is compressed to $ \frac{1}{8} $ of its initial volume, we have: \[ V_2 = \frac{1}{8} V_1 \] Substituting this into the adiabatic equation: \[ P_1 V_1^\gamma = P_2 \left( \frac{1}{8} V_1 \right)^\gamma \] Simplifying: \[ P_1 V_1^\gamma = P_2 \times \left( \frac{1}{8} \right)^\gamma V_1^\gamma \] Canceling $ V_1^\gamma $ from both sides: \[ P_1 = P_2 \times \left( \frac{1}{8} \right)^\gamma \] Since $ \gamma = \frac{5}{3} $, we have: \[ P_1 = P_2 \times \left( \frac{1}{8} \right)^{\frac{5}{3}} \] Now, calculate $ \left( \frac{1}{8} \right)^{\frac{5}{3}} $: \[ \left( \frac{1}{8} \right)^{\frac{5}{3}} = \frac{1}{8^{\frac{5}{3}}} = \frac{1}{32} \] Therefore: \[ P_2 = P_1 \times 32 \] Thus, the ratio of final pressure to initial pressure is: \[ \frac{P_2}{P_1} = 32 \] Therefore, the correct answer is Option (2) — $ 32 $.