The average translational kinetic energy of - nitrogen (molar mass 28) molecules at a particular temperature…
- 0.021
- 0.048
- 0.056
- 0.084
Solution
Translational Kinetic Energy is given by, $\begin{array}{ll} & E=\frac{3}{2} k T \\ \therefore & E \propto T \\ \therefore & \frac{E_{\left(N_2\right)}}{E_{\left(O_2\right)}}=\frac{T_1}{T_2} \\ \therefore & \frac{0.042}{E_{\left(O_2\right)}}=\frac{T}{2 T} \\ \therefore & E_{\left(O_2\right)}=0.084 \mathrm{eV} \end{array}$
Asked in: MHT CET 2024 (02 May Shift 2)
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