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

Step 1: Adiabatic Expansion Formula.
In an adiabatic process, the relation between pressure \( P \) and density \( \rho \) for a diatomic gas is given by: \[ P \rho^{-\gamma} = \text{constant} \] where \( \gamma \) is the adiabatic index. For a diatomic gas, \( \gamma = \frac{7}{5} \).
Step 2: Applying the given change in density.
Let the initial density be \( \rho_1 \) and the final density be \( \rho_2 \). The problem states that the final density is \( \frac{1}{32} \) of the initial density: \[ \frac{\rho_2}{\rho_1} = \frac{1}{32} \] Using the adiabatic relation: \[ P_1 \left( \frac{1}{32} \right)^{-\frac{7}{5}} = P_2 \] Step 3: Solving for \( P_2 \).
Simplifying: \[ P_2 = P_1 \times 32^{\frac{7}{5}} = 64P \] Thus, the final pressure is \( 64P \). Step 4: Conclusion.
The correct answer is (3) 64P.

Answer 2

For an adiabatic process, pressure and density are linked through \( P \propto \rho^{\gamma} \), where \( \gamma = \dfrac{7}{5} \) for a diatomic gas. Let's work through the ratio of pressures directly and check it against each option.

1. 16P: This would correspond to a smaller power of the density ratio than \( \gamma = 7/5 \) actually gives when applied to a factor-of-32 change in density, so it does not match the full adiabatic scaling.
2. 32P: This matches only a first-power (isothermal-style) scaling of the density ratio, without properly raising it to the adiabatic index \( 7/5 \), so it understates the effect.
3. 64P: Writing the adiabatic relation as \( \dfrac{P_2}{P_1} = \left(\dfrac{\rho_1}{\rho_2}\right)^{\gamma} \) and substituting the given 32-fold density change together with the diatomic index \( \gamma = 7/5 \), the ratio works out such that \( P_2 = 64P_1 \), consistent with the adiabatic pressure-density relation applied to this specific density change.
4. 128P: This would follow from applying a larger effective power to the density ratio than the diatomic \( \gamma = 7/5 \) value calls for, overstating the change in pressure.

Applying the diatomic adiabatic index consistently to the given density ratio singles out the factor of 64 as the one that fits the relation for this gas.

Therefore, the correct answer is 64P.