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Calculate the rms speed of an ideal gas...

Calculate the rms speed of an ideal gas molecule of mass `2.99xx10^(-25)` kg at 300k.

Text Solution

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Given, mass of each molecules = `2.99 xx 10^(-26)kg`
temperature T = 300K
We know that, `" "K_(B)=(R)/(N_(A))=(8.314)/(6.023xx10^(23))=1.38xx10^(-23)jk^(-1)`
`therefore " " V_(mn)=sqrt((1.38xx10^(-27)xx300)/(2.99xx10^(-26))=11.77xxsqrt(1000))`
`i.e. " " V_(mn)=372.2ms^(-1).`
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Knowledge Check

  • The average translational energy and the r.m.s. speed of molecules of a sample of oxygen gas at 300 K" are "6 cdot 21 xx 10^(-21)J and 484 ms^(-1) respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour) :

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    B
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    C
    `12 cdot 42 xx10^(-21) J, 968 ms^(-1)`
    D
    `12 cdot 42 xx10^(-21) J, 968 ms^(-1)`

  • Calculate the work done when 1 mol of an ideal gas is compressed reversible from 1 bar to 4 bar at a constant temperature of 300 K.

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    B
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