JEE MAIN - Physics (2002 - No. 36)
At what temperature is the $$r.m.s$$ velocity of a hydrogen molecule equal to that of an oxygen molecule at $${47^ \circ }C?$$
$$80K$$
$$-73$$ $$K$$
$$3$$ $$K$$
$$20$$ $$K$$
ব্যাখ্যা
$${v_{rms}} = $$$$\sqrt {{{RT} \over M}} $$
For $${v_{rms}}$$ to be equal $${{{T_{{H_2}}}} \over {{M_{{H_2}}}}} = {{{T_{{O_2}}}} \over {{M_{{O_2}}}}}$$
Here $${M_{{H_2}}} = 2;\,\,{M_{{O_2}}} = 32;$$
$${T_{{O_2}}} = 47 + 273 = 320K$$
$$\therefore$$ $${{{T_{{H_2}}}} \over 2} = {{320} \over {32}} $$
$$\Rightarrow {T_{{H_2}}} = 20K$$
For $${v_{rms}}$$ to be equal $${{{T_{{H_2}}}} \over {{M_{{H_2}}}}} = {{{T_{{O_2}}}} \over {{M_{{O_2}}}}}$$
Here $${M_{{H_2}}} = 2;\,\,{M_{{O_2}}} = 32;$$
$${T_{{O_2}}} = 47 + 273 = 320K$$
$$\therefore$$ $${{{T_{{H_2}}}} \over 2} = {{320} \over {32}} $$
$$\Rightarrow {T_{{H_2}}} = 20K$$