The mass of a molecule of gas is `4 xx 10^(-30) kg. If 10^(23)` molecules strike the area of 4 square meter with the velocity `10^(7)` m/sec, then what is the pressure exerted on the surface ?
(Assuming perfectly elastic collision and they are hitting perpendicularly)

Answer

To solve the problem, we need to calculate the pressure exerted on a surface by gas molecules striking it. We will follow these steps:

### Step 1: Identify the given values
- Mass of a molecule, \( m = 4 \times 10^{-30} \, \text{kg} \)
- Number of molecules striking per second, \( N = 10^{23} \)
- Velocity of the molecules, \( v = 10^{7} \, \text{m/s} \)
- Area of the surface, \( A = 4 \, \text{m}^2 \)

### Step 2: Calculate the change in momentum for one molecule
The momentum \( p \) of a single molecule is given by:
\[ p = mv \]
When the molecule strikes the surface and rebounds, the change in momentum \( \Delta p \) is:
\[ \Delta p = mv - (-mv) = mv + mv = 2mv \]

### Step 3: Calculate the total change in momentum per second
The total change in momentum per second for \( N \) molecules is:
\[ \text{Total change in momentum} = N \cdot \Delta p = N \cdot 2mv \]
Substituting the values:
\[ \text{Total change in momentum} = N \cdot 2mv = 10^{23} \cdot 2 \cdot (4 \times 10^{-30}) \cdot (10^{7}) \]

### Step 4: Calculate the force exerted on the surface
The force \( F \) exerted on the surface is equal to the rate of change of momentum:
\[ F = N \cdot 2mv \]

### Step 5: Calculate the pressure exerted on the surface
Pressure \( P \) is defined as force per unit area:
\[ P = \frac{F}{A} = \frac{N \cdot 2mv}{A} \]
Substituting the values:
\[ P = \frac{10^{23} \cdot 2 \cdot (4 \times 10^{-30}) \cdot (10^{7})}{4} \]

### Step 6: Simplify the expression
1. Calculate the numerator:
\[ 10^{23} \cdot 2 \cdot (4 \times 10^{-30}) \cdot (10^{7}) = 8 \times 10^{0} = 8 \]
2. Divide by the area:
\[ P = \frac{8}{4} = 2 \, \text{Pascals} \]

### Final Answer
The pressure exerted on the surface is \( 2 \, \text{Pascals} \).

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