Oxygen, nitrogn and helium gas are kept in three identical adiabatic constaners `P, Q` and `R` respectively at equal pressure. When the gases are pushed to half their original volumes. (initial temperature is same)

Options

A. The final temperature in the three containers will be the same.
B. The final pressures in the three containers will be the same.
C. The pressure of oxygen and nitrogen will be the same but that of helium will be different.
D. The temperature of oxygen and nitrogen will be the same but that of helium will be different

Explanation

To solve the problem, we need to analyze the behavior of the three gases (oxygen, nitrogen, and helium) when they are compressed adiabatically to half their original volume. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the Initial Conditions
We have three identical adiabatic containers (P, Q, R) containing oxygen (O2), nitrogen (N2), and helium (He) respectively. All gases are at the same initial pressure (P0), initial volume (V0), and initial temperature (T0).

### Step 2: Applying the Adiabatic Condition
Since the containers are adiabatic, there is no heat exchange with the surroundings. The relationship for an adiabatic process can be described by the equations:
1. \( PV^\gamma = \text{constant} \)
2. \( TV^{\gamma - 1} = \text{constant} \)

Where \( \gamma \) (gamma) is the heat capacity ratio (specific heat at constant pressure to specific heat at constant volume).

### Step 3: Finding Final Pressure
When the volume is reduced to half, the final volume (Vf) becomes:
\[ V_f = \frac{V_0}{2} \]

Using the first equation for pressure:
\[ P_0 V_0^\gamma = P_f \left(\frac{V_0}{2}\right)^\gamma \]

Rearranging gives:
\[ P_f = P_0 \left(2\right)^\gamma \]

This shows that the final pressure depends on \( \gamma \).

### Step 4: Finding Final Temperature
Using the second equation for temperature:
\[ T_0 V_0^{\gamma - 1} = T_f \left(\frac{V_0}{2}\right)^{\gamma - 1} \]

Rearranging gives:
\[ T_f = T_0 \left(2\right)^{\gamma - 1} \]

### Step 5: Identifying Values of Gamma
- For diatomic gases (O2 and N2), \( \gamma = \frac{7}{5} \).
- For monatomic gas (He), \( \gamma = \frac{5}{3} \).

### Step 6: Analyzing Final Pressure and Temperature
1. **Final Pressure**:
- For O2 and N2:
\[ P_f = P_0 \left(2\right)^{\frac{7}{5}} \]
- For He:
\[ P_f = P_0 \left(2\right)^{\frac{5}{3}} \]
- Conclusion: The final pressure of O2 and N2 will be the same, but different from that of He.

2. **Final Temperature**:
- For O2 and N2:
\[ T_f = T_0 \left(2\right)^{\frac{7}{5} - 1} = T_0 \left(2\right)^{\frac{2}{5}} \]
- For He:
\[ T_f = T_0 \left(2\right)^{\frac{5}{3} - 1} = T_0 \left(2\right)^{\frac{2}{3}} \]
- Conclusion: The final temperature of O2 and N2 will be the same, but different from that of He.

### Conclusion
- The final pressure of oxygen and nitrogen will be the same, but that of helium will be different.
- The final temperature of oxygen and nitrogen will be the same, but that of helium will be different.

### Final Answer
The correct options are:
- The pressure of oxygen and nitrogen will be the same, but that of helium will be different.
- The temperature of oxygen and nitrogen will be the same, but that of helium will be different.