`10^(22)` gas molecules each of mass `10^(-26)` M collide with a surface (perpendicular to it) elastically per second over an area `1m^(2)` with a speed `10^(4) ms^(-1)`, the pressure exerted by the gas molecules will be of the order of:

Options

A. `2Nm^(-2)`
B. `4 Nm^(-2)`
C. `6Nm^(-2)`
D. `8 Nm^(-2)`

Explanation

To find the pressure exerted by the gas molecules, we can follow these steps:

### Step 1: Understand the given data
- Number of gas molecules, \( N = 10^{22} \)
- Mass of each gas molecule, \( m = 10^{-26} \, \text{kg} \)
- Area of the surface, \( A = 1 \, \text{m}^2 \)
- Speed of the gas molecules, \( v = 10^{4} \, \text{m/s} \)

### Step 2: Calculate the change in momentum for one molecule
In an elastic collision with a surface, the change in momentum (\( \Delta p \)) for one molecule is given by:
\[ \Delta p = \text{Final momentum} - \text{Initial momentum} = mv - (-mv) = 2mv \]
Substituting the values:
\[ \Delta p = 2 \cdot (10^{-26} \, \text{kg}) \cdot (10^{4} \, \text{m/s}) = 2 \cdot 10^{-26} \cdot 10^{4} = 2 \cdot 10^{-22} \, \text{kg m/s} \]

### Step 3: Calculate the total change in momentum for all molecules
The total change in momentum for \( N \) molecules is:
\[ \Delta p_{\text{total}} = N \cdot \Delta p = 10^{22} \cdot (2 \cdot 10^{-22}) = 2 \, \text{kg m/s} \]

### Step 4: Calculate the force exerted on the surface
The force (\( F \)) exerted on the surface is the rate of change of momentum. Since the molecules collide with the surface every second, the force is:
\[ F = \Delta p_{\text{total}} / \Delta t = 2 \, \text{N} \quad (\text{since } \Delta t = 1 \text{s}) \]

### Step 5: Calculate the pressure exerted on the surface
Pressure (\( P \)) is defined as force per unit area:
\[ P = \frac{F}{A} = \frac{2 \, \text{N}}{1 \, \text{m}^2} = 2 \, \text{N/m}^2 \]

### Conclusion
The pressure exerted by the gas molecules is \( 2 \, \text{N/m}^2 \), which is of the order of \( 2 \, \text{N/m}^2 \).

---