The $R M S$ speed of helium at 400K in $m s^{- 1}$ is:

15.8

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50

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158

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1580

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Solution

The correct option is C $1580$
$T = 400 k$
Molar mass of Helium is = $0.004 k g$

$V_{r m s} = \sqrt{\frac{3 R T}{M}} (R = 8.314 g a s c o n s t a n t)$

$= \sqrt{\frac{3 \times 8.314 \times 400}{0.004}}$

$= 1580 m s^{- 1}$

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The RMS speed of helium at 400K in ms−1 is:

The correct option is C 1580T=400kMolar mass of Helium is =0.004kgVrms=√3RTM(R=8.314gasconstant)=√3×8.314×4000.004=1580ms−1

The $R M S$ speed of helium at 400K in $m s^{- 1}$ is:

The correct option is C $1580$
$T = 400 k$
Molar mass of Helium is = $0.004 k g$

$V_{r m s} = \sqrt{\frac{3 R T}{M}} (R = 8.314 g a s c o n s t a n t)$

$= \sqrt{\frac{3 \times 8.314 \times 400}{0.004}}$

$= 1580 m s^{- 1}$