Question

Two samples of same gas (\(\gamma = \frac{3}{2}\)) have equal volume. If their volumes are doubled by adiabatic and isothermal processes respectively for sample 1 and 2. Their final pressures are now equal. The ratio of initial pressures is:

• A. \(\sqrt{2}\)
• B. \(\frac{1}{\sqrt{2}}\)
• C. \(\frac{3}{\sqrt{2}}\)
• D. \(\frac{\sqrt{2}}{3}\)

Hint:

For problems involving both adiabatic and isothermal processes, understanding how volume changes affect pressure is key to solving for initial conditions.

Answer 1

The Correct Option is A

Let's define the variables:

• Initial pressure and volume of sample 1: \( P_1 \), \( V \)
• Initial pressure and volume of sample 2: \( P_2 \), \( V \)
• Final pressure of both samples: \( P_f \)
• Final volume of sample 1: \( 2V \) (adiabatic)
• Final volume of sample 2: \( 2V \) (isothermal)

Adiabatic Process (Sample 1):

\[ P_1V_1^\gamma = P_fV_f^\gamma \] \[ P_1V^{\frac{3}{2}} = P_f(2V)^{\frac{3}{2}} \] \[ P_1 = P_f \cdot 2^{\frac{3}{2}} \]

Isothermal Process (Sample 2):

\[ P_2V_2 = P_fV_f \] \[ P_2V = P_f(2V) \] \[ P_2 = 2P_f \]

Ratio of Initial Pressures:

\[ \frac{P_1}{P_2} = \frac{P_f \cdot 2^{\frac{3}{2}}}{2P_f} = \frac{2^{\frac{3}{2}}}{2} = 2^{\frac{1}{2}} = \sqrt{2} \]

Therefore, the ratio of initial pressures is \( \sqrt{2} \).

The correct answer is (1) \( \sqrt{2} \).