(D) The kinetic energy of $n$ moles of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$Given values are:$n = 1 \ mol$$T = 27 + 273 = 300 \

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(D) The kinetic energy of $n$ moles of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$Given values are:$n = 1 \ mol$$T = 27 + 273 = 300 \ K$$R = 8.314 \ J \ mol^{-1} \ K^{-1}$Substituting these values into the formula:$KE = \frac{3}{2} \times 1 \times 8.314 \times 300$$KE = 1.5 \times 8.314 \times 300$$KE = 3741.3 \ J$Rounding to the nearest integer, we get $3741 \ J$.

Class 11 Chemistry States of Matter Kinetic molecular theory of gases and Molecular collisions

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The kinetic energy in $J$ of $1$ mole of $N_2$ at $27^{\circ} C$ is: $\left( R = 8.314 \ J \ mol^{-1} \ K^{-1} \right)$

TS EAMCET 2018 All TS EAMCET

A
$2494$
B
$18706$
C
$7482$
D
$3741$

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The kinetic energy in $J$ of $1$ mole of $N_2$ at $27^{\circ} C$ is: $\left( R = 8.314 \ J \ mol^{-1} \ K^{-1} \right)$

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