(B) The formula for total kinetic energy of an ideal gas is $K.E. = \frac{3}{2} nRT$. Given: $n = 0.6 \ mol$,$T = 27 \ ^oC = 27 + 273 = 300 \ K$,and $

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(B) The formula for total kinetic energy of an ideal gas is $K.E. = \frac{3}{2} nRT$. Given: $n = 0.6 \ mol$,$T = 27 \ ^oC = 27 + 273 = 300 \ K$,and $R = 8.314 \ J \ mol^{-1} K^{-1}$. Substituting the values: $K.E. = \frac{3}{2} \times 0.6 \times 8.314 \times 300$. $K.E. = 0.9 \times 8.314 \times 300 = 2244.78 \ J \approx 2245 \ J$.

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

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What is the total kinetic energy of $0.6 \ mol$ of an ideal gas at a temperature of $27 \ ^oC$ in $J$?

A
$1681$
B
$2245$
C
$1122$
D
$2806$

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Kinetic molecular theory of gases and Molecular collisions

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What is the total kinetic energy of $0.6 \ mol$ of an ideal gas at a temperature of $27 \ ^oC$ in $J$?

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