A gas bulb of 1 L capacity contains 2.0 time | vedclass

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(A) Given: Volume $V = 1 \ L = 10^{-3} \ m^3$,Pressure $P = 7.57 	imes 10^3 \ N \ m^{-2}$,Number of molecules $N = 2.0 	imes 10^{21}$.Number of mole

Vedclass · Accepted Answer

$49426$

Vedclass · Answer

(A) Given: Volume $V = 1 \ L = 10^{-3} \ m^3$,Pressure $P = 7.57 	imes 10^3 \ N \ m^{-2}$,Number of molecules $N = 2.0 	imes 10^{21}$.Number of moles $n = \frac{N}{N_A} = \frac{2.0 	imes 10^{21}}{6.022 	imes 10^{23}} \approx 3.321 	imes 10^{-3} \ mol$.Using the ideal gas equation $PV = nRT$,we find the temperature $T = \frac{PV}{nR} = \frac{7.57 	imes 10^3 	imes 10^{-3}}{3.321 	imes 10^{-3} 	imes 8.314} \approx 274.25 \ K$.The molar mass of $N_2$ is $M = 28 \ g \ mol^{-1} = 28 	imes 10^{-3} \ kg \ mol^{-1}$.The root mean square speed $u_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 	imes 8.314 	imes 274.25}{28 	imes 10^{-3}}} \approx 494.26 \ m \ s^{-1}$.Converting to $cm \ s^{-1}$: $494.26 	imes 100 = 49426 \ cm \ s^{-1}$.

$A$ gas bulb of $1 \ L$ capacity contains $2.0 \times 10^{21}$ molecules of nitrogen gas at a pressure of $7.57 \times 10^3 \ N \ m^{-2}$. Calculate the root mean square speed in $cm \ sec^{-1}$. (in $.5$)

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