What is the approximate most probable velocit | vedclass

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(C) The kinetic energy $(KE)$ of one mole of an ideal gas is given by $KE = \frac{3}{2} RT$.From this, we get $RT = \frac{2}{3} KE$.The most probable

Vedclass · Accepted Answer

$\sqrt{155887} \ m \ s^{-1}$

Vedclass · Answer

(C) The kinetic energy $(KE)$ of one mole of an ideal gas is given by $KE = \frac{3}{2} RT$.From this, we get $RT = \frac{2}{3} KE$.The most probable velocity $(v_{mp})$ is given by $v_{mp} = \sqrt{\frac{2RT}{M}}$.Here, $M$ is the molar mass of oxygen $(O_2)$, which is $32 \ g \ mol^{-1} = 0.032 \ kg \ mol^{-1}$.Given $KE = 3741.3 \ J \ mol^{-1}$.Substituting $RT = \frac{2}{3} 	imes 3741.3$ into the velocity formula:$v_{mp} = \sqrt{\frac{2 	imes (\frac{2}{3} 	imes 3741.3)}{0.032}}$$v_{mp} = \sqrt{\frac{4 	imes 3741.3}{3 	imes 0.032}} = \sqrt{\frac{14965.2}{0.096}} = \sqrt{155887.5} \ m \ s^{-1}$.Thus, the approximate most probable velocity is $\sqrt{155887} \ m \ s^{-1}$.Therefore, option $(C)$ is the correct answer.

What is the approximate most probable velocity of oxygen, if the kinetic energy of one mole of oxygen is $3741.3 \ J$?

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