If $$10^{22}$$ gas molecules each of mass $$10^{-26}$$ kg collides with a surface (perpendicular to it) elastically per second over an area 1 m$$^{2}$$ with a speed $$10^{4}$$ m/s, the pressure exerted by the gas molecules will be of the order of:

We have been told that every second $$10^{22}$$ gas molecules strike a perfectly rigid surface of area $$1\ \text{m}^2$$ at right angles. Each molecule has a mass $$m = 10^{-26}\ \text{kg}$$ and a speed $$v = 10^{4}\ \text{m s}^{-1}$$. Because the collisions are stated to be perfectly elastic and normal (i.e. perpendicular), each molecule reverses its velocity on impact.

For one molecule the initial momentum component perpendicular to the wall is $$+mv$$, while the final momentum (after rebounding straight back) is $$-mv$$. The change in momentum for a single collision is therefore

$$\Delta p = p_{\text{final}} - p_{\text{initial}} = (-mv) - (+mv) = -2mv.$$

Only the magnitude matters when calculating force, so the momentum transferred to the wall per collision is

$$|\Delta p| = 2mv.$$

Now, $$N = 10^{22}$$ such collisions occur every second. The total momentum imparted to the wall each second is therefore

$$\text{Total momentum per second} = N \times 2mv = 10^{22} \times 2 \times 10^{-26}\ \text{kg} \times 10^{4}\ \text{m s}^{-1}.$$

Multiplying the powers of ten step by step:

$$10^{22} \times 10^{-26} = 10^{(-26+22)} = 10^{-4},$$

and then

$$10^{-4} \times 10^{4} = 10^{(-4+4)} = 10^{0} = 1.$$

So the numerical product simplifies to

$$N \times 2mv = 2 \times 1 = 2\ \text{kg m s}^{-2}.$$

But $$1\ \text{kg m s}^{-2} = 1\ \text{N}$$, hence the force exerted on the wall is

$$F = 2\ \text{N}.$$

By definition, pressure is force per unit area. Stating the formula first, we write

$$P = \frac{F}{A}.$$

Since $$A = 1\ \text{m}^2$$, we obtain

$$P = \frac{2\ \text{N}}{1\ \text{m}^2} = 2\ \text{Pa}.$$

Hence, the correct answer is Option A.