TS EAMCET · Chemistry · States of Matter
The average kinetic energy of one molecule of an ideal gas at \(27^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure is
- A \(900 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\)
- B \(6.21 \times 10^{-21} \mathrm{JK}^{-1}\) molecule \(^{-1}\)
- C 336.7 \(\mathrm{JK}^{-1}\) molecule \(^{-1}\)
- D \(3741.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(B) \(6.21 \times 10^{-21} \mathrm{JK}^{-1}\) molecule \(^{-1}\)
Step-by-step Solution
Detailed explanation
Average kinetic energy per molecule \(=\frac{3}{2} k T\) or \(\quad=\frac{3}{2} \frac{R}{N_0} T\) \(\begin{aligned} & =\frac{3}{2} \times \frac{8.314}{6.023 \times 10^{23}} \times 300 \\ & =6.21 \times 10^{-21} \mathrm{JK}^{-1} \text { molecule }^{-1}\end{aligned}\)
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