The average velocity of molecules of a gas of molecilar weight (M) at temperature (T) is

Options

A. `sqrt (3 R T)/(M)`.
B. `sqrt (8 R T)/(pi M)`.
C. `sqrt (2 R T)/(M)`.
D. zero

Answer

To find the average velocity of molecules of a gas with molecular weight \( M \) at temperature \( T \), we can use the kinetic theory of gases. Here is a step-by-step solution:

### Step 1: Understand the formula for average velocity
The average velocity \( V_{\text{avg}} \) of gas molecules is given by the formula:
\[ V_{\text{avg}} = \sqrt{\frac{8kT}{\pi m}} \]
where:
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature,
- \( m \) is the mass of a single molecule of the gas.

### Step 2: Relate molecular mass to molar mass
The molecular weight \( M \) is the mass of one mole of the gas, which can also be expressed in terms of the mass of a single molecule \( m \) and Avogadro's number \( N_A \):
\[ M = m \cdot N_A \]
Thus, we can express \( m \) as:
\[ m = \frac{M}{N_A} \]

### Step 3: Substitute \( m \) in the average velocity formula
Substituting the expression for \( m \) into the average velocity formula gives:
\[ V_{\text{avg}} = \sqrt{\frac{8kT}{\pi \left(\frac{M}{N_A}\right)}} \]
This simplifies to:
\[ V_{\text{avg}} = \sqrt{\frac{8kT \cdot N_A}{\pi M}} \]

### Step 4: Substitute Boltzmann constant \( k \)
We can relate the Boltzmann constant \( k \) to the universal gas constant \( R \):
\[ k = \frac{R}{N_A} \]
Substituting this into the equation, we get:
\[ V_{\text{avg}} = \sqrt{\frac{8 \left(\frac{R}{N_A}\right) T \cdot N_A}{\pi M}} \]
This simplifies to:
\[ V_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}} \]

### Final Answer
Thus, the average velocity of the molecules of a gas with molecular weight \( M \) at temperature \( T \) is:
\[ V_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}} \]