Question 179: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

Momentum imparted to the surface in one collision,
$$\Delta p = ({p_i} - {p_f}) = mv - ( - mv) = 2mv$$ ..... (i)
Force on the surface due to n collision per second, $$F = {n \over t}(\Delta p) = n\Delta p$$ ($$\because$$ $$t = 1s$$)
= 2 mnv [from Eq. (i)]
So, pressure on the surface,
$$p = {F \over A} = {{2mnv} \over A}$$
Here, m = 10
$$-$$26
kg, n = 10
22
s
$$-$$1
, v = 10
4
ms
$$-$$1
, A = 1 m
2
$$	herefore$$ Pressure, $$p = {{2 	imes {{10}^{ - 26}} 	imes {{10}^{22}} 	imes {{10}^4}} \over 1} = 2$$ N/m
2
So, pressure exerted is of order of 10
0
.

Answer

108 N/m2

Answer

1016 N/m2

Answer

104 N/m2

Answer

2 N/m2

Centre of Mass and Rotational Motion

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

Choose the correct answer: