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

Options

A. `10^(4)N//m^(2)`
B. `10^(8)N//m^(2)`
C. `10^(3)N//m^(2)`
D. `10^(16)N//m^(2)`

Answer

To find the pressure exerted by the gas molecules, we can use the formula for pressure in terms of momentum transfer due to collisions. The pressure \( P \) can be calculated using the formula:

\[ P = \frac{F}{A} \]

where \( F \) is the force exerted by the gas molecules on the surface, and \( A \) is the area over which the force is applied.

### Step 1: Calculate the change in momentum per collision

When a gas molecule collides elastically with a surface, it reverses its velocity. The change in momentum \( \Delta p \) for one molecule is given by:

\[ \Delta p = m(v_f - v_i) \]

where:
- \( m = 10^{-26} \, \text{kg} \) (mass of one molecule),
- \( v_f = -100 \, \text{m/s} \) (final velocity, negative because it reverses direction),
- \( v_i = 100 \, \text{m/s} \) (initial velocity).

Thus, the change in momentum is:

\[ \Delta p = 10^{-26} \, \text{kg} \times (-100 - 100) \, \text{m/s} = 10^{-26} \, \text{kg} \times (-200) \, \text{m/s} = -2 \times 10^{-24} \, \text{kg m/s} \]

### Step 2: Calculate the total momentum change per second

The total number of molecules colliding with the surface per second is \( N = 10^{22} \). Therefore, the total change in momentum per second (which is equal to the force \( F \)) is:

\[ F = N \times \Delta p = 10^{22} \times (-2 \times 10^{-24}) = -2 \times 10^{-2} \, \text{N} \]

The negative sign indicates the direction of the force, but we are interested in the magnitude:

\[ F = 2 \times 10^{-2} \, \text{N} \]

### Step 3: Calculate the pressure

The area \( A \) is given as \( 1 \, \text{m}^2 \). Thus, the pressure \( P \) is:

\[ P = \frac{F}{A} = \frac{2 \times 10^{-2} \, \text{N}}{1 \, \text{m}^2} = 2 \times 10^{-2} \, \text{Pa} \]

### Step 4: Express the pressure in terms of order of magnitude

The pressure exerted by the gas molecules is of the order of:

\[ P \approx 2 \times 10^{-2} \, \text{Pa} \implies P \text{ is of the order of } 10^{-2} \, \text{Pa} \]

### Final Answer

The pressure exerted by the gas molecules will be of the order of \( 10^{-2} \, \text{Pa} \).

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