According to kinetic theory of gases ,
A . The motion of the gas molecules freezes at `0^(@)C`
B. The mean free path of gas molecules decreases if the density of molecules is increased .
C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant .
D . Average kinetic energy per molecule per degree of freedon is `3/2k_(B)T` (for monoatomic gases ).
Choose the most appropriate answer from the options given below :

Options

A. A and C only
B. B and C only
C. A and B only
D. C and D only

Explanation

To solve the question based on the kinetic theory of gases, we will analyze each statement provided in the options.

**Step 1: Analyze Option A**
- The first statement claims that the motion of gas molecules freezes at 0°C.
- According to kinetic theory, molecular motion ceases at absolute zero, which is 0 Kelvin (-273.15°C), not at 0°C.
- Therefore, **Option A is incorrect**.

**Step 2: Analyze Option B**
- The second statement suggests that the mean free path of gas molecules decreases if the density of molecules increases.
- The mean free path (λ) is given by the formula:
\[ \lambda = \frac{kT}{\sqrt{2} \pi d^2 p} \]
where \( p \) is pressure, \( d \) is the diameter of the molecules, and \( T \) is the temperature.
- If the density increases, the pressure also increases (for a constant temperature), which leads to a decrease in the mean free path.
- Thus, **Option B is correct**.

**Step 3: Analyze Option C**
- The third statement claims that the mean free path of gas molecules increases if the temperature is increased while keeping the pressure constant.
- From the mean free path formula, if the temperature \( T \) increases while keeping pressure constant, the mean free path \( \lambda \) will indeed increase since it is directly proportional to temperature.
- Therefore, **Option C is also correct**.

**Step 4: Analyze Option D**
- The fourth statement asserts that the average kinetic energy per molecule per degree of freedom is \( \frac{3}{2} k_B T \) for monoatomic gases.
- However, this expression represents the total average kinetic energy for three degrees of freedom (translational motion) of a monoatomic gas. The average kinetic energy per degree of freedom is actually \( \frac{1}{2} k_B T \).
- Thus, **Option D is incorrect**.

**Conclusion:**
- The correct options are B and C. Since the question asks for the most appropriate answer, we can conclude that **Option B and C are correct**, but if only one answer is to be chosen, we can select **Option B** as it directly relates to the density and mean free path relationship.