The average kinetic energy of one molecule of an ideal gas at $27^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$…
The average kinetic energy of one molecule of an ideal gas at $27^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure is
- $900 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$
- $6.21 \times 10^{-21} \mathrm{JK}^{-1}$ molecule $^{-1}$
- 336.7 $\mathrm{JK}^{-1}$ molecule $^{-1}$
- $3741.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
Solution
Average kinetic energy per molecule
$=\frac{3}{2} k T$
or $\quad=\frac{3}{2} \frac{R}{N_0} T$
$\begin{aligned} & =\frac{3}{2} \times \frac{8.314}{6.023 \times 10^{23}} \times 300 \\ & =6.21 \times 10^{-21} \mathrm{JK}^{-1} \text { molecule }^{-1}\end{aligned}$
Asked in: AP EAMCET 2009
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