According to the kinetic theory of gases, the r.m.s velocity of gas molecules is directly proportional to :

T

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\sqrt{T}

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T^{2}

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\frac{1}{\sqrt{T}}

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Solution

The correct option is B $\sqrt{T}$
According to the kinetic theory of gases
$v_{r m s} = \sqrt{\frac{3 R T}{M}}$ (standard result)
so $v_{r m s} α \sqrt{T}$
answer is B.

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Q. Use kinetic theory of gases to show that the average Kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

According to the kinetic theory of gases, in an ideal gas between two successive collisions, a gas molecule travels.....

According to pressure equation, PV=1/3 M Vrms².

=>nRT=1/3 M Vrms²

=> T directly proportional to M Vrms²

According to mean Kinetic energy,

KE(avg)=1/2 M Vrms²

=> KE(avg) directly proportional to M Vrms²

This means.. M Vrms² directly proportional to KE(avg) T

=> M Vrms² /T = KE(avg) × proportionality constant

So this means the avg Kinetic energy is inversely proportional to Absolute Temperature. But actually KE(avg) is directly proportional to T. Please clear my doubt

Q. According to kinetic theory of gases, the collisions between molecules of a gas:

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According to the kinetic theory of gases, the r.m.s velocity of gas molecules is directly proportional to :

The correct option is B √TAccording to the kinetic theory of gasesvrms=√3RTM (standard result)so vrmsα√Tanswer is B.

The correct option is B $\sqrt{T}$
According to the kinetic theory of gases
$v_{r m s} = \sqrt{\frac{3 R T}{M}}$ (standard result)
so $v_{r m s} α \sqrt{T}$
answer is B.