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) \( \sqrt{2} \).

Answer 2

To solve the problem, we need to calculate the ratio of the initial pressures of two gas samples with the same gas constant and volume, where the volumes are doubled by adiabatic and isothermal processes, respectively.

1. Understanding the Given Information:
We are given two gas samples with the same gas constant and initial volume. The final pressures of the two samples are equal, and we need to find the ratio of their initial pressures. The volume of the first sample is doubled by an adiabatic process, while the volume of the second sample is doubled by an isothermal process.

2. Analyzing the Adiabatic Process:
For an adiabatic process, the relationship between pressure and volume is given by:

\[ P_1 V_1^\gamma = P_2 V_2^\gamma \] where \( P_1 \) and \( P_2 \) are the initial and final pressures, \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( \gamma = \frac{3}{2} \) is the adiabatic index. Since the volume is doubled, we have: \[ P_1 V_1^{3/2} = P_2 V_2^{3/2} \] Since the final pressures are equal, we can simplify: \[ P_1 = P_2 \left(\frac{V_2}{V_1}\right)^{3/2} = P_2 (2)^{3/2} = P_2 \sqrt{8} \] Thus, the initial pressure \( P_1 \) is \( \sqrt{8} \) times the final pressure \( P_2 \).

3. Analyzing the Isothermal Process:
For an isothermal process, the relationship between pressure and volume is given by:

\[ P_1 V_1 = P_2 V_2 \] Since the volume is doubled, we have: \[ P_1 = P_2 \frac{V_2}{V_1} = P_2 \cdot 2 \] Thus, the initial pressure \( P_1 \) is \( 2 \) times the final pressure \( P_2 \).

4. Finding the Ratio of Initial Pressures:
We are asked to find the ratio of the initial pressures of the two gas samples. Using the results from the adiabatic and isothermal processes:

\[ \frac{P_1^{\text{adiabatic}}}{P_1^{\text{isothermal}}} = \frac{\sqrt{8}}{2} = \sqrt{2} \]

5. Conclusion:
The ratio of the initial pressures is \( \sqrt{2} \).

Final Answer:
The correct answer is (B) \( \sqrt{2} \).