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A fresh air is composed of ntirogen N(2)...

A fresh air is composed of ntirogen `N_(2)(78%)` and oxygen `O_(2)(21%)`. Find the rms speed of `N_(2)andO_(2)` at `20^(@)C`.

Text Solution

Verified by Experts

For nitrogen,
Molar mass m = 0.0280 kg/mol
Temperature `T=20^(@)C=20+273=293K`
`R=8.314Jmol^(-1)k^(-1)`
`V_(rms)=sqrt((3RT)/(m))`
`=sqrt((3xx8.31xx293)/(0.0280))=sqrt(2610xx10^(2))`
`=511ms^(-1)`
`V_(rms)" of "N_(2)=511ms^(-1)`
For `O_(2)`
Molar mass `M=0.0320kgmol^(-1)`
`V_(rms)=sqrt((3xx8.31xx293)/(0.0320))=sqrt(2280xx10^(2))`
`V_(rms)" of "O_(2)=478ms^(-1)`
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A fresh air is composed of nitrogen N_(2)(78%) and oxygen O_(2)(21%) . Find the rms speed of N_(2) and O_(2) at 20^(@) C.

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Knowledge Check

  • Assertion : N_(2) is more stable than O_(2) . Reason : Bond order of N_(2) is more than that of O_(2) .

    A
    Both assertion and reason are true and reason is the correct explanation of the assertion
    B
    Both assertion and reason are true and reason is not the correct explanation of the assertion.
    C
    Assertion is true but reason is false
    D
    Both assertion and reason are false.

  • For the reaction N_(2)O(g)rarr2NO_(2)(g)+1/2O_(2)(g) , the value of rate of disappearance of N_(2)O_(5) is given as 6.5xx10^(-2)"mol L"^(-1)s^(-1) . The rate of formation of NO_(2) and O_(2) is given respectively as :

    A
    `(3.25xx10^(-2)"mol" L^(-1)s^(-1))` and `(1.3xx10^(-2)"mol"L^(-1)s^(-1))`
    B
    `(1.3xx10^(-2)"mol"L^(-1)s^(-1))` and `(3.25xx10^(-2)"mol"L^(-1)s^(-1))`
    C
    `(1.3xx10^(-1)"mol"L^( -1)s^(-1))` and `(3.25xx10^(-2)"mol"L^(-1)s^(-1))`
    D
    None of these

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