A container of volume 2.24 , L can withstand | vedclass

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(D) Using the ideal gas equation $pV = nRT$,where $p = 2 \, atm$,$V = 2.24 \, L$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 298

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$4.2$

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(D) Using the ideal gas equation $pV = nRT$,where $p = 2 \, atm$,$V = 2.24 \, L$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 298 \, K$.$n = \frac{pV}{RT} = \frac{2 	imes 2.24}{0.0821 	imes 298} \approx 0.183 \, mol$.The molar mass of $N_2$ is $28 \, g/mol$.The maximum mass of $N_2$ is $m = n 	imes M = 0.183 	imes 28 \approx 5.124 \, g$.Since the container explodes at $2 \, atm$,the safe amount must be slightly less than $5.124 \, g$. Among the given options,$4.2 \, g$ is the closest value that ensures safety.

$A$ container of volume $2.24 \, L$ can withstand a maximum pressure of $2 \, atm$ at $298 \, K$ before exploding. The maximum amount of nitrogen (in $g$) that can be safely put in this container at this temperature is closest to

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