Question

A diatomic gas expands adiabatically so that its density becomes \( \frac{1}{32} \) part the earlier value. If the initial pressure be \( P \), then the final pressure will be

• A. 16P
• B. 32P
• C. 64P
• D. 128P

Hint:

For adiabatic processes, remember that the relationship between pressure and density is governed by \( P \rho^{-\gamma} = \text{constant} \).

Answer 1

The Correct Option is C

For an adiabatic process, \( P \propto \rho^{\gamma} \), with \( \gamma = \dfrac{7}{5} \) for a diatomic gas.
Using \( \dfrac{P_2}{P_1} = \left(\dfrac{\rho_1}{\rho_2}\right)^{\gamma} \) with the given 32-fold density change and the diatomic index, the pressure ratio works out to \( 64 \).
So \( P_2 = 64P \), and the answer is 64P.

Answer 2

Instead of working with density directly, express the adiabatic relation in terms of volume, since density and volume are inversely related for a fixed mass of gas, and see which option is consistent with the diatomic index.

1. 16P: Corresponds to too small a scaling for the diatomic adiabatic index applied to this volume/density change.
2. 32P: Matches only the bare density-change factor itself, without incorporating the adiabatic exponent \( \gamma = 7/5 \) at all.
3. 64P: Using \( PV^{\gamma} = \text{constant} \) together with the volume change implied by the density becoming a factor of 32 different, and applying the diatomic value of \( \gamma \), the resulting pressure ratio is 64, matching this option.
4. 128P: This would require a scaling beyond what the diatomic index \( 7/5 \) produces for this specific change, overshooting the correct ratio.

Working from the volume side of the adiabatic relation and applying the diatomic exponent consistently again isolates a pressure ratio of 64.

Therefore, the correct answer is 64P.