JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 2)
For a given gas at 1 atm pressure, rms speed
of the molecule is 200 m/s at 127°C. At 2 atm
pressure and at 227°C, the rms speed of the
molecules will be :
100 m/s
100 $$\sqrt 5 $$ m/s
80 $$\sqrt 5 $$ m/s
80 m/s
Explanation
$${V_{rms}} = \sqrt {{{3RT} \over {{M_w}}}} $$
$$ \Rightarrow {V_{rms}} \propto \sqrt T $$
Now, $${v \over {200}} = \sqrt {{{500} \over {400}}} $$
$$ \Rightarrow {v \over {200}} = {{\sqrt 5 } \over 2}$$
$$ \Rightarrow v = 100\sqrt 5 $$ m/s
$$ \Rightarrow {V_{rms}} \propto \sqrt T $$
Now, $${v \over {200}} = \sqrt {{{500} \over {400}}} $$
$$ \Rightarrow {v \over {200}} = {{\sqrt 5 } \over 2}$$
$$ \Rightarrow v = 100\sqrt 5 $$ m/s
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