Question

A monoatomic gas is compressed to \((1/8)\) th of its volume adiabatically. If its initial pressure is \(P_{0}\), then increase in the pressure will be

• A. \(32\ P_{0}\)
• B. \(31\ P_{0}\)
• C. \(16\ P_{0}\)
• D. \(15\ P_{0}\)

Hint:

Use \(PV^{\gamma}=\text{constant}\) with \(\gamma=5/3\) for a monoatomic gas.

Answer 1

The Correct Option is B

Step 1: Formula-first approach
$P_{0}V^{5/3}=P_{2}\left(\dfrac{V}{8}\right)^{5/3}$, so $P_{2}=P_{0}\,8^{5/3}$.

Step 2: Simplify
$8^{5/3}=(8^{1/3})^{5}=2^{5}=32$.

Step 3: Why the other options fail
$32P_{0}$ is the final pressure, not the increase. $16P_{0}$ and $15P_{0}$ come from using $\gamma=\dfrac{4}{3}$, the value for a polyatomic gas, not a monoatomic one.

Final Answer:
The rise in pressure is 31 P0. \[ \boxed{\text{(B) }31\ P_{0}} \]