At what temperature will be total kinetic energy (KE) of $0.30$ mole of $H e$ be the same as the total KE of $0.40$ mole of $A r$ at $400 K$ :

400 K

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373 K

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533 K

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300 K

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Solution

The correct option is A $533 K$
$K E = \frac{3 n R T}{2}$
$\frac{(K E) H e}{(K E) A r} = \frac{(m o l e s o f H e) \times (t e m p o f H e)}{(m o l e s o f A r) \times (t e m p r a t u r e o f A r)}$
$1 = \frac{0.3 \times T}{0.4 \times 400}$
$T = 400 \times \frac{4}{3}$
$T = 533 K$
So, option $C$ is correct

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