Kinetic energy in kJ of 280 g of N_2 at 27^{ | vedclass

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(B) Number of moles of $N_2$ $(n)$ = $\frac{	ext{mass}}{	ext{molar mass}} = \frac{280 \ g}{28 \ g \ mol^{-1}} = 10 \ mol$. Temperature $(T)$ = $27^{

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$37.4$

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(B) Number of moles of $N_2$ $(n)$ = $\frac{	ext{mass}}{	ext{molar mass}} = \frac{280 \ g}{28 \ g \ mol^{-1}} = 10 \ mol$. Temperature $(T)$ = $27^{\circ} C = (27 + 273) \ K = 300 \ K$. The formula for kinetic energy of an ideal gas is $K.E. = \frac{3}{2} nRT$. Substituting the values: $K.E. = \frac{3}{2} 	imes 10 \ mol 	imes 8.314 \ J \ mol^{-1} \ K^{-1} 	imes 300 \ K$. $K.E. = 1.5 	imes 10 	imes 8.314 	imes 300 = 37413 \ J$. Converting to $kJ$: $37413 \ J = 37.413 \ kJ \approx 37.4 \ kJ$.

Kinetic energy in $kJ$ of $280 \ g$ of $N_2$ at $27^{\circ} C$ is approximately $(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

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