The kinetic energy of 2 , text{moles} of N_2 | vedclass

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Question

(C) The formula for the kinetic energy of an ideal gas is given by $KE = \frac{3}{2} nRT$. Here,$n = 2 \, 	ext{moles}$,$R = 8.314 \, J \, K^{-1} \, m

Vedclass · Accepted Answer

$7491.6$

Vedclass · Answer

(C) The formula for the kinetic energy of an ideal gas is given by $KE = \frac{3}{2} nRT$. Here,$n = 2 \, 	ext{moles}$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$,and $T = 27 + 273 = 300 \, K$. Substituting the values: $KE = \frac{3}{2} 	imes 2 	imes 8.314 	imes 300$. $KE = 3 	imes 8.314 	imes 300 = 7482.6 \, J$. Rounding to the nearest provided option,the correct answer is $7491.6 \, J$.

The kinetic energy of $2 \, \text{moles}$ of $N_2$ gas at $27 \, ^oC$ is .............. $J$.

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