JEE MAIN - Physics (2018 (Offline) - No. 13)
The mass of a hydrogen molecule is 3.32 $$\times$$ 10-27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45o to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly:
2.35 $$\times$$ 103 N m-2
4.70 $$\times$$ 103 N m-2
2.35 $$\times$$ 102 N m-2
4.70 $$\times$$ 102 N m-2
Explanation
Considering one hydrogen molecule :
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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
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
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