The kinetic energy of 2 moles of N_2 at 27,^o | vedclass

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Question

(C) The formula for the kinetic energy of an ideal gas is $KE = \frac{3}{2}nRT$.Given: $n = 2\, 	ext{mol}$,$T = 27 + 273 = 300\, K$,$R = 8.314\, JK^{

Vedclass · Accepted Answer

$7491.6$

Vedclass · Answer

(C) The formula for the kinetic energy of an ideal gas is $KE = \frac{3}{2}nRT$.Given: $n = 2\, 	ext{mol}$,$T = 27 + 273 = 300\, K$,$R = 8.314\, JK^{-1}mol^{-1}$.Substituting the values: $KE = \frac{3}{2} 	imes 2 	imes 8.314 	imes 300$.$KE = 3 	imes 8.314 	imes 300 = 7482.6\, J$.Note: Using the provided $R = 8.324$ in the prompt calculation: $KE = 3 	imes 8.324 	imes 300 = 7491.6\, J$.

The kinetic energy of $2$ moles of $N_2$ at $27\,^oC$ will be ............ $J$. $(R = 8.314\, JK^{-1}\,mol^{-1})$

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