MHT CET · Physics · Kinetic Theory of Gases
The average translational kinetic energy of - nitrogen (molar mass 28) molecules at a particular temperature is 0.042 eV . The translational kinetic energy of oxygen molecules (molar mass 32 ) in eV at double the temperature is
- A 0.021
- B 0.048
- C 0.056
- D 0.084
Answer & Solution
Correct Answer
(D) 0.084
Step-by-step Solution
Detailed explanation
\(\mathrm{E}_{\left(\mathrm{N}_2\right)}=0.042 \mathrm{eV}\)
Translational Kinetic Energy is given by,
\(\begin{array}{ll}
& E=\frac{3}{2} k T \\
\therefore & E \propto T \\
\therefore & \frac{E_{\left(N_2\right)}}{E_{\left(O_2\right)}}=\frac{T_1}{T_2} \\
\therefore & \frac{0.042}{E_{\left(O_2\right)}}=\frac{T}{2 T} \\
\therefore & E_{\left(O_2\right)}=0.084 \mathrm{eV}
\end{array}\)
Translational Kinetic Energy is given by,
\(\begin{array}{ll}
& E=\frac{3}{2} k T \\
\therefore & E \propto T \\
\therefore & \frac{E_{\left(N_2\right)}}{E_{\left(O_2\right)}}=\frac{T_1}{T_2} \\
\therefore & \frac{0.042}{E_{\left(O_2\right)}}=\frac{T}{2 T} \\
\therefore & E_{\left(O_2\right)}=0.084 \mathrm{eV}
\end{array}\)
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