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.