The RMS speed of helium gas molecules at 127- C isA- 1000 m - sB- 1228 m - sC- 1570 m - sD- 1470 m - s

The correct option is C 1570 nbsp;m/sV_rms = √(3RT/M) = √(3 × 8.3 × 400/4 × 10^ 3) = √(3 × 8.3 × 10^5) = √(25 × 10^5) = 5 × 100 × 3.14 = 314 × 5 = 1570 m/s ...

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Standard XII

Chemistry

Introduction

The RMS speed...

Question

The RMS speed of helium gas molecules at $127^{\circ} C$ is

1000

m / s

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1228

m / s

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1570

m / s

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1470

m / s

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Solution

The correct option is C 1570 $m / s$
$V_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 \times 8.3 \times 400}{4 \times 10^{- 3}}}$
$= \sqrt{3 \times 8.3 \times 10^{5}} = \sqrt{25 \times 10^{5}} = 5 \times 100 \times 3.14$
$= 314 \times 5 = 1570 m / s$

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The RMS speed of helium gas molecules at 127 ∘C is

The correct option is C 1570 m/sVrms=√3RTM=√3×8.3×4004×10−3 =√3×8.3×105=√25×105=5×100×3.14 =314×5=1570 m/s

The RMS speed of helium gas molecules at $127^{\circ} C$ is

The correct option is C 1570 $m / s$
$V_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 \times 8.3 \times 400}{4 \times 10^{- 3}}}$
$= \sqrt{3 \times 8.3 \times 10^{5}} = \sqrt{25 \times 10^{5}} = 5 \times 100 \times 3.14$
$= 314 \times 5 = 1570 m / s$