JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 13)
For a particular ideal gas which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?
_28th_January_Morning_Shift_en_13_1.png)
_28th_January_Morning_Shift_en_13_2.png)
_28th_January_Morning_Shift_en_13_3.png)
_28th_January_Morning_Shift_en_13_4.png)
Explanation
We know, for an ideal gas,
$${V_{rms}} = \sqrt {{{3RT} \over M}} $$
$$ \Rightarrow {V_{ms}} = V_{rms}^2 = {{3RT} \over M}$$
$$ \Rightarrow {V_{ms}} \propto T$$
$$ \Rightarrow {V_{ms}} \propto T$$
i.e. mean squared velocity is directly proportional to temperature.
$$ \Rightarrow {V_{ms}} = {\left( {{{3R} \over M}} \right)^T}$$
by comparing it with $y=mx$ (straight line)
_28th_January_Morning_Shift_en_13_5.png)
Hence, option 3 is correct.
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