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Standard XII

Chemistry

Fluorides of Xenon

A mixture of ...

Question

A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds $(\frac{v_{r m s} (h e l i u m)}{v_{r m s} (a r g o n)})$ is :

0.32

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0.45

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2.24

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3.16

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Solution

The correct option is B 3.16
$V_{r m s} = \sqrt{\frac{3 R T}{M}}$
Thus we get required ratio as $= \sqrt{\frac{M_{A r}}{M_{H e}}} = \sqrt{10}$
$= 3.16$

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Similar questions

A mixture of 2 moles of heìium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds

(\frac{v_{r m s} (h e l i u m)}{v_{r m s} (a r g o n)})

Q. A mixture of

2

moles of helium gas (atomic mass

= 4 u

), and

1

mole of argon gas (atomic mass

= 40 u

) is kept at

300 K

in a container. The ratio of their rms speeds

[\frac{V_{r m s} (h e l i u m)}{V_{r m s} (a r g o n)}]

, is close to:

Q. A mixture of

2

moles of helium gas

(atomic mass = 4u)

, and

1

mole of argon gas

(atomic mass = 40u)

is kept at

300 K

in a container. The ratio of their

R M S

speeds

[\frac{v_{rms} (helium)}{v_{rms} (argon)}]

is close to:

Q. Temperature at which rms speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at

- 20^{\circ} C

is
(Take molar mass of argon as

40 g

and for helium as

4 g

)

Q. Temperature at which rms speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at

- 20^{\circ} C

is
(Take molar mass of argon as

40 g

and for helium as

4 g

)

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A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds (vrms(helium)vrms(argon)) is :

The correct option is B 3.16Vrms=√3RTMThus we get required ratio as =√MArMHe=√10=3.16

A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds $(\frac{v_{r m s} (h e l i u m)}{v_{r m s} (a r g o n)})$ is :

The correct option is B 3.16
$V_{r m s} = \sqrt{\frac{3 R T}{M}}$
Thus we get required ratio as $= \sqrt{\frac{M_{A r}}{M_{H e}}} = \sqrt{10}$
$= 3.16$