If 1022 gas molecules each of mass 10-26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m 2 with a speed 104 m / s, the pressure exerted by the gas molecules will be of the order of

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 m 2 with a speed 104 m / s, the pressure exerted by the gas molecules will be of the order of (A) 104 N / m 2 (B) 108 N / m 2 (C) 103 N / m 2 (D) None of the above

Tardigrade Admin · Accepted Answer

Momentum imparted to the surface in one collision,

Δ p = (p_{i} - p_{f}) = m v - (- m v) = 2 m

Force on the surface due to

n

collision per second,

F = \frac{n}{t} (Δ p) = n Δ p (∵ t = 1 s)

= 2 mn v f ro m Eq . (i)]

So, pressure on the surface,

p = \frac{F}{A} = \frac{2 mn v}{A}

Here,

m = 1 0^{- 26} k g, n = 1 0^{22} s^{- 1}, v = 1 0^{4} m s^{- 1}

,

A = 1 m^{2}

∴

Pressure,

p = \frac{2 \times 1 0 ^{- 26} \times 1 0 ^{22} \times 1 0 ^{4}}{1} = 2 N / m^{2}

So, pressure exerted is of order of

1 0^{\circ}

.

Q. If $1 0^{22}$ gas molecules each of mass $1 0^{- 26} k g$ collide with a surface (perpendicular to it) elastically per second over an area $1 m^{2}$ with a speed $1 0^{4} m / s$ , the pressure exerted by the gas molecules will be of the order of

$1 0^{4} N / m^{2}$

$1 0^{8} N / m^{2}$

$1 0^{3} N / m^{2}$

None of the above

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

104N/m2

108N/m2

103N/m2

None of the above

Q. If $1 0^{22}$ gas molecules each of mass $1 0^{- 26} k g$ collide with a surface (perpendicular to it) elastically per second over an area $1 m^{2}$ with a speed $1 0^{4} m / s$ , the pressure exerted by the gas molecules will be of the order of

$1 0^{4} N / m^{2}$

$1 0^{8} N / m^{2}$

$1 0^{3} N / m^{2}$