The total kinetic energy of 1 text{ mole} of | vedclass

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Question

(B) The total kinetic energy $E_T$ of an ideal gas is given by the formula: $E_T = \frac{f}{2} nRT$Where $n$ is the number of moles,$f$ is the degrees

Vedclass · Accepted Answer

$1500 	ext{ cal}$

Vedclass · Answer

(B) The total kinetic energy $E_T$ of an ideal gas is given by the formula: $E_T = \frac{f}{2} nRT$Where $n$ is the number of moles,$f$ is the degrees of freedom,$R$ is the universal gas constant,and $T$ is the temperature in Kelvin.Given: $n = 1 	ext{ mole}$,$T = 27^{\circ}C = 27 + 273 = 300 	ext{ K}$,$R = 2 	ext{ cal/mol K}$.For a diatomic gas like $N_2$,the degrees of freedom $f = 5$ (at room temperature).Substituting the values: $E_T = \frac{5}{2} 	imes 1 	imes 2 	imes 300$$E_T = 5 	imes 300 = 1500 	ext{ cal}$.

The total kinetic energy of $1 \text{ mole}$ of $N_2$ at $27^{\circ}C$ is approximately: $(R = 2 \text{ cal/mol K})$

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