JEE Main · PhysicsMedium

$A container of volume V contains a mixture of two ideal gases, one of molar mass M_{1} and the other of molar mass M_{2}, in equal number of moles. If the temperature of the mixture is T, what is the root mean square speed of the gas molecules in the mixture?$

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Correct answer: C — $\frac{6 R T}{M _{1} + M _{2}}$

Solution

$The number of moles of each gas is equal, so the average molar mass is M_{a v g} = \frac{M _{1} + M _{2}}{2} .$
$The root mean square speed is given by v_{r m s} = \frac{3 R T}{M _{a v g}} .$
$Substituting M_{a v g} into the formula gives v_{r m s} = \frac{3 R T}{( M _{1} + M _{2} ) /2} = \frac{6 R T}{M _{1} + M _{2}} .$
$Hence the answer is (C).$