Statement-1: In equation `P=(1)/(3)alphaV_(rms)^(2)`, the term `alpha ` represents density of gas. ltBrgt Statement-2: `V_(rms)=sqrt((3RT)/(M))`
Statement-3: Rotational kinetic energy of a monoatomic gas is zero.

Options

A. FFF
B. TTT
C. TFF
D. FFT

Answer

To analyze the statements provided in the question, we will evaluate each statement one by one.

### Step-by-Step Solution:

**Statement 1**: In the equation \( P = \frac{1}{3} \alpha V_{rms}^2 \), the term \( \alpha \) represents the density of gas.

1. We know that the pressure \( P \) of an ideal gas can be expressed as:
\[ P = \frac{1}{3} \rho V_{rms}^2 \]
where \( \rho \) is the density of the gas and \( V_{rms} \) is the root mean square velocity of the gas molecules.

2. Given the equation \( P = \frac{1}{3} \alpha V_{rms}^2 \), we can compare it with the standard equation.

3. From the comparison, it is clear that \( \alpha \) must equal \( \rho \) (the density of the gas).

4. Therefore, **Statement 1 is True**.

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**Statement 2**: \( V_{rms} = \sqrt{\frac{3RT}{M}} \)

1. The root mean square velocity \( V_{rms} \) for an ideal gas can be derived from the kinetic theory of gases.

2. It is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas.

3. This formula is well-established in thermodynamics and kinetic theory.

4. Therefore, **Statement 2 is True**.

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**Statement 3**: The rotational kinetic energy of a monoatomic gas is zero.

1. A monoatomic gas consists of single atoms (e.g., noble gases like helium, neon).

2. Monoatomic gases do not have rotational degrees of freedom because they do not have a structure that allows for rotation.

3. The kinetic energy associated with translational motion can be calculated using the degrees of freedom. For a monoatomic gas, the degrees of freedom are:
- Translational: 3 (in x, y, z directions)
- Rotational: 0

4. Since rotational kinetic energy is related to the rotational degrees of freedom, and there are none for a monoatomic gas, the rotational kinetic energy is indeed zero.

5. Therefore, **Statement 3 is True**.

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### Conclusion:
All three statements are true. Thus, the answer is:
- Statement 1: True
- Statement 2: True
- Statement 3: True

### Final Answer:
**All statements are true (TTT)**.

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