Find the kinetic energy of 1 g of nitrogen a | vedclass

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(A) The molar mass of nitrogen $(N_2)$ is $M = 28 \ g/mol$.The temperature in Kelvin is $T = 77 + 273 = 350 \ K$.The kinetic energy $(KE)$ of an ideal

Vedclass · Accepted Answer

$155.8$

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(A) The molar mass of nitrogen $(N_2)$ is $M = 28 \ g/mol$.The temperature in Kelvin is $T = 77 + 273 = 350 \ K$.The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = \frac{3}{2} nRT$,where $n$ is the number of moles.The number of moles $n = \frac{	ext{mass}}{Molar \ mass} = \frac{1 \ g}{28 \ g/mol} = \frac{1}{28} \ mol$.Substituting the values into the formula:$KE = \frac{3}{2} 	imes \left( \frac{1}{28} ight) 	imes 8.31 	imes 350$.$KE = \frac{3 	imes 8.31 	imes 350}{56} = \frac{8725.5}{56} \approx 155.8 \ J$.

Find the kinetic energy of $1 \ g$ of nitrogen at $77^{\circ} \ C$. $(R = 8.31 \ J \ mol^{-1} K^{-1})$ (in $J$)

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