Three identical adiabatic containers `A, B and C` Contain helium, neon and oxygen respectively at equal pressure. The gases are pushed to half their original volumes. (initial temperature is same)

Options

A. The final temperature of the gas in each container is same
B. The final pressure of the gas in each container is same
C. The final temperature of both helium and neon is same
D. The final pressure of both helium and neon is same

Explanation

To solve the problem step by step, we will analyze the adiabatic process for the three gases (helium, neon, and oxygen) in identical containers.

### Step 1: Understand the Adiabatic Process
In an adiabatic process, there is no heat exchange with the surroundings. The relationship between pressure (P), volume (V), and temperature (T) is governed by the adiabatic index (γ).

### Step 2: Use the Adiabatic Relation for Pressure
For an adiabatic process, we have the relation:
\[ P_i V_i^\gamma = P_f V_f^\gamma \]
where \( P_i \) and \( V_i \) are the initial pressure and volume, and \( P_f \) and \( V_f \) are the final pressure and volume.

### Step 3: Substitute the Final Volume
Since the final volume is half of the initial volume:
\[ V_f = \frac{V_i}{2} \]
Substituting this into the adiabatic relation gives:
\[ P_i V_i^\gamma = P_f \left(\frac{V_i}{2}\right)^\gamma \]

### Step 4: Rearranging for Final Pressure
Rearranging the equation:
\[ P_f = P_i \cdot 2^\gamma \]
This shows that the final pressure depends on the adiabatic index (γ).

### Step 5: Determine the Values of γ for Each Gas
- For helium (monatomic gas): \( \gamma = \frac{5}{3} \)
- For neon (monatomic gas): \( \gamma = \frac{5}{3} \)
- For oxygen (diatomic gas): \( \gamma = \frac{7}{5} \)

### Step 6: Compare Final Pressures
From the equation \( P_f = P_i \cdot 2^\gamma \):
- Final pressure for helium and neon will be the same because they have the same γ.
- Final pressure for oxygen will be higher due to a larger γ.

### Step 7: Use the Adiabatic Relation for Temperature
For temperature, we use the relation:
\[ T_i V_i^{\gamma - 1} = T_f V_f^{\gamma - 1} \]
Substituting \( V_f = \frac{V_i}{2} \):
\[ T_i V_i^{\gamma - 1} = T_f \left(\frac{V_i}{2}\right)^{\gamma - 1} \]

### Step 8: Rearranging for Final Temperature
Rearranging gives:
\[ T_f = T_i \cdot 2^{\gamma - 1} \]

### Step 9: Compare Final Temperatures
- For helium and neon (both have \( \gamma = \frac{5}{3} \)):
\[ T_f = T_i \cdot 2^{\frac{5}{3} - 1} = T_i \cdot 2^{\frac{2}{3}} \]

- For oxygen (\( \gamma = \frac{7}{5} \)):
\[ T_f = T_i \cdot 2^{\frac{7}{5} - 1} = T_i \cdot 2^{\frac{2}{5}} \]

### Step 10: Conclusion
- Final pressure: \( P_f \) for helium and neon is the same, but \( P_f \) for oxygen is higher.
- Final temperature: \( T_f \) for helium and neon is the same, but \( T_f \) for oxygen is higher.

### Final Answers
1. Final pressure of each gas in the container is NOT the same.
2. Final pressure of helium and neon is the same.
3. Final temperature of the gases in each container is NOT the same.
4. Final temperature of helium and neon is the same.