Pick the correct statement(s):

Options

A. The rms translational speed for different ideal gas samples at the same temperature is not necessarily same but it depends on the molar mass
B. Speed of a particular gas molecule can be greater the RMS speed for the sample.
C. Temperature of an ideal gas is doubled from `100^@ C` to `200^@ C`. The average kinetic energy of each particle is also doubled
D. It is possible for both the pressure and volume of a monoatomic ideal gas to change simultaneously without causing the internal energy of the gas to change

Explanation

To solve the question, we need to evaluate each of the four statements provided and determine which ones are correct based on our understanding of the gaseous state and thermodynamics.

### Step-by-Step Solution:

**Step 1: Evaluate Statement A**
- Statement A claims that the RMS (Root Mean Square) translational speed for different ideal gas samples at the same temperature is not necessarily the same but depends on the molecular molar mass.
- The formula for RMS speed is given by:
\[ v_{RMS} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas.
- Since \( M \) varies for different gases, the RMS speed will also vary, even if the temperature is the same.
- **Conclusion**: Statement A is **correct**.

**Step 2: Evaluate Statement B**
- Statement B states that the speed of a particular gas molecule can be greater than the RMS speed of the sample.
- The RMS speed is a statistical measure and does not represent the speed of any individual molecule. For example, if we have molecules with speeds of 1 m/s, 2 m/s, and 3 m/s, the RMS speed can be calculated as:
\[ v_{RMS} = \sqrt{\frac{1^2 + 2^2 + 3^2}{3}} = \sqrt{\frac{1 + 4 + 9}{3}} = \sqrt{4} = 2 \text{ m/s} \]
- Here, the molecule with a speed of 3 m/s is indeed greater than the calculated RMS speed of 2 m/s.
- **Conclusion**: Statement B is **correct**.

**Step 3: Evaluate Statement C**
- Statement C claims that if the temperature of an ideal gas is doubled from 100°C to 200°C, the average kinetic energy of each particle is also doubled.
- The average kinetic energy \( KE \) of a gas particle is given by:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin.
- Converting temperatures:
- \( T_1 = 100°C = 373 K \)
- \( T_2 = 200°C = 473 K \)
- The change in temperature in Kelvin is not a simple doubling (it’s an increase of 100 K, not a doubling of energy).
- Therefore, the average kinetic energy does not double.
- **Conclusion**: Statement C is **incorrect**.

**Step 4: Evaluate Statement D**
- Statement D states that it is possible for both the pressure and volume of a monoatomic ideal gas to change simultaneously without causing the internal energy of the gas to change.
- The internal energy \( U \) of an ideal gas depends only on its temperature. If the temperature remains constant, the internal energy does not change.
- According to the ideal gas law \( PV = nRT \), if pressure increases while volume decreases (or vice versa), the temperature can remain constant, thus keeping the internal energy constant.
- **Conclusion**: Statement D is **correct**.

### Final Conclusion:
The correct statements are **A, B, and D**.