Question:

A polyatomic gas ($\gamma = 4/3$) is compressed to $\left(\frac{1}{8}\right)^{\text{th}}$ of its volume adiabatically. If its initial pressure is $P_0$, its new pressure will be

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For adiabatic volume compressions, if the volume drops by a factor of $x$, the pressure increases by a factor of $x^\gamma$. Here, the compression factor is 8 and $\gamma = 4/3$. Taking the cube root of 8 first gives 2, and then raising 2 to the $4^{\text{th}}$ power yields 16 instantly. Splitting the fractional exponent into root-then-power makes the calculation simple to perform mentally!
Updated On: Jun 18, 2026
  • $2P_0$
  • $8P_0$
  • $6P_0$
  • $16P_0$
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Question:
The problem presents an ideal polyatomic gas undergoing an adiabatic compression process. The volume is reduced to an eighth of its initial value, and we need to determine the final pressure of the gas system in terms of its initial pressure $P_0$.

Step 2: Key Formula or Approach:

For an adiabatic process, the relationship between pressure ($P$) and volume ($V$) is governed by Poisson's law: $$P V^\gamma = \text{constant}$$ This can be written for initial and final states as: $$P_1 V_1^\gamma = P_2 V_2^\gamma \implies P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma$$

Step 3: Detailed Explanation:

Let's list the given parameters: Initial pressure, $P_1 = P_0$ Initial volume, $V_1 = V$ Final volume, $V_2 = \frac{V}{8}$ Adiabatic index, $\gamma = \frac{4}{3}$ Substitute these values into our state equation to isolate the final pressure $P_2$: $$P_2 = P_0 \left(\frac{V}{\frac{V}{8}}\right)^{4/3}$$ The volume variables cancel out completely: $$P_2 = P_0 (8)^{4/3}$$ We can rewrite the number 8 as a base power of 2, i.e., $8 = 2^3$: $$P_2 = P_0 \left(2^3\right)^{4/3}$$ Applying the laws of exponents, multiply the power indices together: $$P_2 = P_0 \cdot 2^{3 \times \frac{4}{3}} = P_0 \cdot 2^4$$ Evaluating $2^4 = 16$ gives: $$P_2 = 16P_0$$

Step 4: Final Answer:

The new pressure of the compressed gas is $16P_0$, which corresponds to option (D).
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