Q17 / 120 · VITEEE 2009 · Chemistry
The average kinetic energy of one molecule of an ideal gas at $27^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure is
- A900 cal $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$
- Correct answerB$6.21 \times 10^{-21} \mathrm{JK}^{-1}$ molecule $^{-1}$
- C$336.7 \mathrm{JK}^{-1}$ molecules $^{-1}$
- D$3741.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
Solution
Average kinetic energy per molecule
$$
=\frac{3}{2} k T
$$
or $\frac{3}{2} \frac{R}{N_{0}} T=\frac{3}{2} \times \frac{8.314}{6.023 \times 10^{23}} \times 300$ $=6.21 \times 10^{-21} \mathrm{JK}^{-1}$ molecule $^{-1}$
Hence, the correct answer is option: 'B'.