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Question

(B) The average kinetic energy per molecule of an ideal gas is given by the formula: $KE_{avg} = \frac{3}{2} k T$. Here, $k$ is the Boltzmann constant

Vedclass · Accepted Answer

$6.21 	imes 10^{-21} \ J \ 	ext{molecule}^{-1}$

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(B) The average kinetic energy per molecule of an ideal gas is given by the formula: $KE_{avg} = \frac{3}{2} k T$. Here, $k$ is the Boltzmann constant $(k = \frac{R}{N_A})$, $T$ is the temperature in Kelvin, and $N_A$ is Avogadro's number. Given: $T = 27^{\circ} C = 27 + 273 = 300 \ K$. Substituting the values: $KE_{avg} = \frac{3}{2} 	imes \left( \frac{8.314 \ J \ K^{-1} \ mol^{-1}}{6.022 	imes 10^{23} \ mol^{-1}} ight) 	imes 300 \ K$. $KE_{avg} = 1.5 	imes 1.38 	imes 10^{-23} \ J \ K^{-1} 	imes 300 \ K$. $KE_{avg} \approx 6.21 	imes 10^{-21} \ J \ 	ext{molecule}^{-1}$.

The average kinetic energy of one molecule of an ideal gas at $27^{\circ} C$ and $1 \ atm$ pressure is

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