If the kinetic energy and RMS speed of a gas | vedclass

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(B) The kinetic energy $(KE)$ of one mole of an ideal gas is given by the formula: $KE = \frac{1}{2} M v_{rms}^2$,where $M$ is the molar mass in $kg \

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$32$

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(B) The kinetic energy $(KE)$ of one mole of an ideal gas is given by the formula: $KE = \frac{1}{2} M v_{rms}^2$,where $M$ is the molar mass in $kg \ mol^{-1}$ and $v_{rms}$ is the root mean square speed in $m \ s^{-1}$.Given: $KE = 4.0 \ kJ \ mol^{-1} = 4000 \ J \ mol^{-1}$ and $v_{rms} = 5.0 	imes 10^4 \ cm \ s^{-1} = 500 \ m \ s^{-1}$.Substituting the values into the equation: $4000 = \frac{1}{2} 	imes M 	imes (500)^2$.$4000 = \frac{1}{2} 	imes M 	imes 250000$.$4000 = M 	imes 125000$.$M = \frac{4000}{125000} = 0.032 \ kg \ mol^{-1}$.Converting to grams: $M = 0.032 	imes 1000 \ g \ mol^{-1} = 32 \ g \ mol^{-1}$.

If the kinetic energy and $RMS$ speed of a gas at a certain temperature are $4.0 \ kJ \ mol^{-1}$ and $5.0 \times 10^4 \ cm \ s^{-1}$,respectively,the molecular weight of the gas is:

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