Question

The pressure \(P_1\) and density \(d_1\) of a diatomic gas change to \(P_2\) and \(d_2\) during an adiabatic operation. Find the value of \[ \frac{P_1}{P_2}, \] if \[ \frac{d_2}{d_1}=32. \]

• A. \(128\)
• B. \[ \frac{1}{64} \]
• C. \(64\)
• D. \[ \frac{1}{128} \]

Hint:

For adiabatic processes, \[ P\propto d^\gamma. \] Remember: \[ \gamma=\frac53 \text{ (monoatomic)}, \qquad \gamma=\frac75 \text{ (diatomic)}. \] Convert the density ratio directly into a pressure ratio using this relation.

Answer 1

The Correct Option is D

Concept: For an adiabatic process, \[ PV^\gamma=\text{constant}. \] Since density \[ d=\frac{m}{V}, \] we have \[ V\propto\frac1d. \] Hence, the adiabatic relation can also be written as \[ P\propto d^\gamma. \] For a diatomic gas, \[ \gamma=\frac75. \]

Step 1: Write the adiabatic relation in terms of density. \[ \frac{P_2}{P_1} = \left(\frac{d_2}{d_1}\right)^\gamma. \] Given, \[ \frac{d_2}{d_1}=32. \] Therefore, \[ \frac{P_2}{P_1} = 32^{7/5}. \]

Step 2: Evaluate the power. Since \[ 32=2^5, \] \[ 32^{7/5} = (2^5)^{7/5} = 2^7 = 128. \] Thus, \[ \frac{P_2}{P_1}=128. \]

Step 3: Find \(\frac{P_1}{P_2}\). \[ \frac{P_1}{P_2} = \frac1{128}. \] Therefore, \[ \boxed{\frac{P_1}{P_2}=\frac1{128}} \] \[ \boxed{\text{Answer = (D)}} \]