If the RMS speed of nitrogen at a certain tem | vedclass

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(B) Given,$U_{rms} = 3000 \ ms^{-1}$.The formula for $RMS$ speed is $U_{rms} = \sqrt{\frac{3RT}{M}}$.Squaring both sides,we get $U_{rms}^2 = \frac{3RT

Vedclass · Accepted Answer

$126$

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(B) Given,$U_{rms} = 3000 \ ms^{-1}$.The formula for $RMS$ speed is $U_{rms} = \sqrt{\frac{3RT}{M}}$.Squaring both sides,we get $U_{rms}^2 = \frac{3RT}{M}$.For nitrogen $(N_2)$,the molar mass $M = 28 \ g/mol = 28 	imes 10^{-3} \ kg/mol$.Substituting the values: $(3000)^2 = \frac{3RT}{28 	imes 10^{-3}}$.$9 	imes 10^6 = \frac{3RT}{28 	imes 10^{-3}}$.$3RT = 9 	imes 10^6 	imes 28 	imes 10^{-3} = 252 	imes 10^3 \ J/mol$.The kinetic energy $(K.E.)$ for $1$ mole of an ideal gas is given by $K.E. = \frac{3}{2} RT$.$K.E. = \frac{1}{2} 	imes (3RT) = \frac{1}{2} 	imes 252 	imes 10^3 \ J/mol = 126 	imes 10^3 \ J/mol = 126 \ kJ$.

If the $RMS$ speed of nitrogen at a certain temperature is $3000 \ ms^{-1}$,the approximate kinetic energy of one mole of nitrogen at that temperature in $kJ$ is (assume nitrogen as ideal gas)

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