The total kinetic energy of 0.6 mol of an id | vedclass

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Question

(B) The total kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$. Given: $n = 0.6 \ mol$,$T = 27 \ ^oC

Vedclass · Accepted Answer

$2245$

Vedclass · Answer

(B) The total kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$. Given: $n = 0.6 \ mol$,$T = 27 \ ^oC = 27 + 273 = 300 \ K$,and $R = 8.314 \ J \ mol^{-1} K^{-1}$. Substituting the values: $KE = \frac{3}{2} 	imes 0.6 	imes 8.314 	imes 300$. $KE = 1.5 	imes 0.6 	imes 8.314 	imes 300 = 0.9 	imes 8.314 	imes 300 = 2244.78 \ J$. Rounding to the nearest integer,we get $2245 \ J$.

The total kinetic energy of $0.6 \ mol$ of an ideal gas at $27 \ ^oC$ is .......... $J$.

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