The total kinetic energy of 1 mole of oxygen | vedclass

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(C) The total kinetic energy of an ideal gas is given by the formula: $K.E. = \frac{f}{2} nRT$.For a diatomic gas like oxygen $(O_2)$,the degrees of f

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$6232.5$

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(C) The total kinetic energy of an ideal gas is given by the formula: $K.E. = \frac{f}{2} nRT$.For a diatomic gas like oxygen $(O_2)$,the degrees of freedom $(f)$ is $5$ at room temperature.Given:Number of moles $(n)$ = $1 \ mol$Temperature $(T)$ = $27^{\circ} C = 27 + 273 = 300 \ K$Universal gas constant $(R)$ = $8.31 \ J/mol \cdot K$Substituting the values into the formula:$K.E. = \frac{5}{2} 	imes 1 	imes 8.31 	imes 300$$K.E. = 5 	imes 8.31 	imes 150$$K.E. = 6232.5 \ J$

The total kinetic energy of $1$ mole of oxygen at $27^{\circ} C$ is:
[Use universal gas constant $(R) = 8.31 \ J/mol \cdot K$] (in $J$)

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The total kinetic energy of $1$ mole of oxygen at $27^{\circ} C$ is:[Use universal gas constant $(R) = 8.31 \ J/mol \cdot K$] (in $J$)