QJEEM19Phd3dd: JEE Main 2019 Physics - Class 11 JEEM19Phd3dd Solution: If 10 22 gas molecules each of mass 10 –26 kg collide with a…

Momentum imparted to the surface in one collision,

$$\mathrm{\Delta }p=(p_i-p_f)=mv-(-mv)=2mv$$ ..... (i)

Force on the surface due to n collision per second, $$F=\frac{n}{t}(\mathrm{\Delta }p)=n\mathrm{\Delta }p$$ ($$\because$$ $$t=1s$$)

= 2 mnv [from Eq. (i)]

So, pressure on the surface,

$$p=\frac{F}{A}=\frac{2mnv}{A}$$

Here, m = 10^$$-$$26 kg, n = 10²² s^$$-$$1, v = 10⁴ ms^$$-$$1, A = 1 m²

$$\therefore$$ Pressure, $$p=\frac{2\times {10}^{-26}\times {10}^{22}\times {10}^4}{1}=2$$ N/m²

So, pressure exerted is of order of 10⁰.