QJEEM18Phb176: JEE Main 2018 Physics - Class 11 JEEM18Phb176 Solution: The mass of a hydrogen molecule is 3.32 10 -27 kg. If 10 23…

Considering one hydrogen molecule :



As collision is elastic so, e = 1

Initial momentum,

$\overrightarrow {{P_i}}$ = ${{mv} \over {\sqrt 2 }}$ $\widehat i$ $-$ ${{mv} \over {\sqrt 2 }}$ $\widehat j$

Final momentum,

$\overrightarrow {{P_f}}$ = ${{mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$ $-$ ${{mv} \over {\sqrt 2 }}\widehat j$

$\therefore\,\,\,$ Change in momentum for single H molecule,

$\Delta$ P = $\overrightarrow {{P_f}}$ $-$ $\overrightarrow {{P_i}}$

= ${{2mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$

$\therefore\,\,\,$ $\left| {\Delta P} \right|$ = ${{2mv} \over {\sqrt 2 }}$

Now for n hydrogen molecule total momentum changes per second,

= $\left( {{{2mv} \over {\sqrt 2 }}} \right)$ $\times$ n

As we know,

Force (F) = ${{\Delta P} \over {\Delta t}}$

= ${{{2mv} \over {\sqrt 2 }}}$ $\times$ n

$\therefore\,\,\,$ As direction of $\Delta$ P is towards negative i so force on the molecule will also be towards negative i direction. From Newton's third law, the reaction force will be on the wall in positive i direction with same magnitude.

$\therefore\,\,\,$ Force on the wall = ${{{2mv} \over {\sqrt 2 }}}$ $\times$ n

$\therefore\,\,\,$ Pressure on the wall, P = ${F \over A}$

= ${{2mv\,n} \over {\sqrt 2 A}}$

= ${{2 \times 3.32 \times {{10}^{ - 27}} \times {{10}^3} \times {{10}^{23}}} \over {\sqrt 2 \times2\times {{10}^{ - 4}}}}$

= 2.35 $\times$ 10³ N m ^{$-$ 2}