Question

If 10²² gas molecules each of mass 10^-26 kg collide with a surface (perpendicular to it) elastically per second over an area of 1 m² with a speed of 10⁴ m/s, the pressure exerted by the gas molecules will be of the order of ______.

विकल्प

• 10⁴ N/m²

• 10³ N/m²

• 10⁸ N/m²

• 10¹⁶ N/m²

Answer 1

If 10²² gas molecules each of mass 10^-26 kg collide with a surface (perpendicular to it) elastically per second over an area of 1 m² with a speed of 10⁴ m/s, the pressure exerted by the gas molecules will be of the order of 10⁴ N/m².

Explanation:

Given: The number of gas molecules is n = 10²², their mass is m = 10^-26 kg each, and their speed is v = 10⁴ m/s. The area of the surface over which the gas molecules are colliding elastically every second is A = 1 m².

To find: The order of the pressure exerted on the surface by the gas molecules.

The change in momentum of a gas molecule colliding with the surface elastically is

Δp = mv - (-mv) = 2mv

When n gas molecules collide with a surface in an elastic collision, their momentum changes.

Δp = mv - (-mv) = 2nmv .......(i)

The surface force exerted by n gas molecules:

F = `(Δp)/(Δt) = (2nmv)/1` = 2nmv ......(ii) (Δt = 1s)

The surface pressure exerted by n gas molecules:

P = `F/A = (2mnv)/A`

= `(2 xx 10^-26 xx 10^22 xx 10^4)/1 = 2`

As a result, the order of pressure exerted on the surface by gas molecules is 0.