Heat And Thermodynamics Ques 4
4 If $10^{22}$ gas molecules each of mass $10^{-26} \mathrm{~kg}$ collide with a surface (perpendicular to it) elastically per second over an area $1 \mathrm{~m}^2$ with a speed $10^4 \mathrm{~m} / \mathrm{s}$, the pressure exerted by the gas molecules will be of the order of
(2019 Main, 8 April I)
(a) $10^4 \mathrm{~N} / \mathrm{m}^2$
(b) $10^8 \mathrm{~N} / \mathrm{m}^2$
(c) $10^3 \mathrm{~N} / \mathrm{m}^2$
(d) $10^{16} \mathrm{~N} / \mathrm{m}^2$
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Answer:
Correct Answer: 4.( * )
Solution:
Formula:
- Momentum imparted to the surface in one collision,
$ \Delta p=\left(p_i-p_f\right)=m v-(-m v)=2 m v $
Force on the surface due to $n$ collisions per second,
$ \begin{aligned} F & =\frac{n}{t}(\Delta p)=\frac{n}{t} \Delta p \\ & = 2 m n v \quad [from Eq. (i)] \end{aligned} $
So, pressure on the surface, increases with depth.
$ p=\frac{F}{A}=\frac{2 m n v}{A} $
Here, $m=10^{-26} \mathrm{~kg}, n=10^{22} \mathrm{~s}^{-1}$,
$v=10^4 \mathrm{~ms}^{-1}, A=1 \mathrm{~m}^2$
$\therefore$ Pressure, $\quad p=\frac{2 \times 10^{-26} \times 10^{22} \times 10^4}{1}=2 \mathrm{~Pa}$
So, pressure exerted is order of $10^{\circ}\text{C}$ or $10^{\circ}\text{F}$.
BETA
