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Molecular Speeds - Vrms, Vavg, Vmps

The temperatu...

Question

The temperature at which the $R M S$ velocity of $C H_{4}$ is equal to the $R M S$ velocity of helium at $27^{o} C$ is:

927 K

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927^{o} C

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1200^{o} C

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623^{o} C

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Solution

The correct option is B $927^{o} C$
When $R M S$ velocities are equal,
$\frac{T_{1}}{T_{2}} = \frac{M_{1}}{M_{2}}$
$\frac{T_{1}}{300} = \frac{16}{4}$
$T_{1} = 1200$ $K$
Thus, $T_{1} = 927^{0} C$

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