Assertion Prssure of a gas is `2/3` times translational kinetic energy of gas molecules.
ReasonTranslational degree of freedom of any type of gas is three, whether the gas is monoatomic, diatomic or polyatomic.

Options

A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B. If both Assertion and Reason are correct but Reason is not the correct explation os Assertion.
C. If Assertionis true bur Reason is false.
D. If Assertion is false but Reason is true.

Explanation

To analyze the given assertion and reason, we can break down the solution into the following steps:

### Step 1: Understanding the Assertion
The assertion states that the pressure of a gas is \( \frac{2}{3} \) times the translational kinetic energy of gas molecules.

**Hint:** Recall the relationship between pressure and kinetic energy in gases.

### Step 2: Translational Kinetic Energy of Gas Molecules
The translational kinetic energy (TKE) of gas molecules can be expressed as:
\[ TKE = \frac{1}{2} m v_{rms}^2 \]
where \( v_{rms} \) is the root mean square velocity of the gas molecules.

**Hint:** Remember that \( v_{rms} \) can be derived from the ideal gas law.

### Step 3: Expression for \( v_{rms} \)
From the kinetic theory of gases, the root mean square velocity is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant and \( M \) is the molar mass of the gas.

**Hint:** Substitute \( v_{rms} \) back into the TKE equation.

### Step 4: Substitute \( v_{rms} \) into TKE
Substituting \( v_{rms} \) into the TKE equation gives:
\[ TKE = \frac{1}{2} m \left(\sqrt{\frac{3RT}{M}}\right)^2 = \frac{1}{2} m \cdot \frac{3RT}{M} = \frac{3}{2} \cdot \frac{mRT}{M} \]
Here, \( n = \frac{m}{M} \) (number of moles), so:
\[ TKE = \frac{3}{2} nRT \]

**Hint:** Relate TKE to the ideal gas equation.

### Step 5: Ideal Gas Equation
From the ideal gas equation, we have:
\[ PV = nRT \]
Thus, we can express \( nRT \) as:
\[ nRT = PV \]

**Hint:** Substitute \( nRT \) back into the TKE expression.

### Step 6: Relate TKE to Pressure
Substituting \( nRT \) into the TKE equation gives:
\[ TKE = \frac{3}{2} PV \]
Now, if we consider the kinetic energy per unit volume, we can express it as:
\[ \text{Kinetic Energy per unit volume} = \frac{TKE}{V} = \frac{3}{2} P \]

**Hint:** Compare the derived expression with the assertion.

### Step 7: Conclusion on Assertion
From the above, we can conclude that:
\[ P = \frac{2}{3} \cdot \text{Kinetic Energy per unit volume} \]
This confirms that the assertion is correct.

**Hint:** Now, check the reason provided.

### Step 8: Understanding the Reason
The reason states that the translational degree of freedom of any type of gas is three, regardless of whether it is monoatomic, diatomic, or polyatomic.

**Hint:** Recall the definition of degrees of freedom in the context of gas molecules.

### Step 9: Degrees of Freedom
For translational motion, all types of gases (monoatomic, diatomic, polyatomic) have three degrees of freedom associated with motion along the x, y, and z axes.

**Hint:** This supports the reason being correct.

### Step 10: Final Conclusion
Both the assertion and the reason are correct, and the reason correctly explains the assertion.

**Final Answer:** Both the assertion and reason are true, and the reason is the correct explanation for the assertion.