The kinetic energy of 1 mole of He gas at 27^ | vedclass

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(B) The kinetic energy of $1$ mole of an ideal gas is given by the formula $K.E. = \frac{3}{2}RT$.Here,$R$ is the universal gas constant,$R = 8.314 \,

Vedclass · Accepted Answer

$3735$

Vedclass · Answer

(B) The kinetic energy of $1$ mole of an ideal gas is given by the formula $K.E. = \frac{3}{2}RT$.Here,$R$ is the universal gas constant,$R = 8.314 \, J \cdot mol^{-1} \cdot K^{-1}$ (often approximated as $8.3 \, J \cdot mol^{-1} \cdot K^{-1}$ in calculations).The temperature $T$ in Kelvin is $T = 27 + 273 = 300 \, K$.Substituting the values: $K.E. = \frac{3}{2} 	imes 8.3 	imes 300$.$K.E. = 1.5 	imes 8.3 	imes 300 = 3735 \, J$.

The kinetic energy of $1$ mole of $He$ gas at $27^{\circ}C$ is .... $J$.

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