Pressure exerted by one mole of an ideal gas kept in a vessel of 'V' L having root mean square speed of molecules 'v' and 'm' mass of each molecule is correctly given by the equation

Options

A. `P=(1)/(2)(N_(A))/(V)mv^(2)`
B. `P=(1)/(3)(N_(A))/(V)mv^(2)`
C. `P=(2)/(3)(N_(A))/(V)mv^(2)`
D. `P=(3)/(2)(N_(A))/(V)mv^(2)`

Answer

To derive the pressure exerted by one mole of an ideal gas in a vessel of volume \( V \) liters, with a root mean square speed \( v \) and mass \( m \) of each molecule, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relation of RMS Speed**:
The root mean square (RMS) speed \( v_{\text{RMS}} \) of gas molecules is given by the formula:
\[ v_{\text{RMS}} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

2. **Square the RMS Speed Equation**:
To eliminate the square root, we square both sides:
\[ v_{\text{RMS}}^2 = \frac{3RT}{M} \]

3. **Rearranging for RT**:
From the squared equation, we can express \( RT \):
\[ RT = \frac{Mv_{\text{RMS}}^2}{3} \]

4. **Using the Ideal Gas Law**:
The ideal gas law is given by:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)):
\[ PV = RT \]

5. **Substituting RT in the Ideal Gas Law**:
We substitute the expression for \( RT \) from step 3 into the ideal gas law:
\[ PV = \frac{Mv_{\text{RMS}}^2}{3} \]

6. **Solving for Pressure \( P \)**:
Rearranging the equation to solve for pressure \( P \):
\[ P = \frac{Mv_{\text{RMS}}^2}{3V} \]

7. **Final Expression**:
Thus, the pressure exerted by one mole of an ideal gas in a vessel of volume \( V \) with root mean square speed \( v \) and mass \( m \) of each molecule is:
\[ P = \frac{1}{3} \frac{N_A m v^2}{V} \]
where \( N_A \) is Avogadro's number (the number of molecules in one mole).

### Final Answer:
\[ P = \frac{1}{3} \frac{N_A m v^2}{V} \]