Q103 / 125 · BITSAT 2012 · Physics
The average translational kinetic energy of $\mathrm{O}_{2}$ (molar mass 32) molecules at a particular temperature is $0.048 \mathrm{eV}$. The translational kinetic energy of $\mathrm{N}_{2}$ (molar mass 28 ) molecules in $\mathrm{eV}$ at the same temperature is
- A0.0015
- B0.003
- Correct answerC0.048
- D0.768
Solution
Given,
\(\mathrm{E}_{\mathrm{O}_{2}}=0.048 \mathrm{eV}\)
We know that,
Translational kinetic energy,
\(\mathrm{E}=\frac{3}{2} \mathrm{KT}\)
where, \(\mathrm{k}=\) Boltzmann constant
\(\mathrm{T}=\) Temperature
\(\therefore \mathrm{E} \propto \mathrm{T}\)
Since, the temperature is same for both oxygen and nitrogen
therefore, \(\mathrm{E}_{\mathrm{O}_{2}}=\mathrm{E}_{\mathrm{N}_{2}}\)
\(\therefore \mathrm{E}_{\mathrm{N}_{2}}=0.048 \mathrm{eV}\)
Hence, the correct answer is option: 'C'.