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:

• A. 28
• B. 32
• C. 40
• D. 16

Hint:

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 $\gamma$.

Answer 1

The Correct Option is B

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) — $32$.