KCET · Physics · Magnetic Effects of Current
The mean energy of a molecule of an ideal gas is
- A \(2 \mathrm{KT}\)
- B \( \frac{3}{2} \mathrm{KT} \)
- C KT
- D \( \frac{1}{2} \mathrm{KT} \)
Answer & Solution
Correct Answer
(B) \( \frac{3}{2} \mathrm{KT} \)
Step-by-step Solution
Detailed explanation
The mean energy of a molecule of an ideal gas depends on its temperature. It is described as
\( K=\frac{3}{2} k_{B} T \)
where \( k_{B} \) is Boltzmann's constant; \( \mathrm{T} \) is temperature; \( \mathrm{K} \) is mean energy per molecule of gas.
\( K=\frac{3}{2} k_{B} T \)
where \( k_{B} \) is Boltzmann's constant; \( \mathrm{T} \) is temperature; \( \mathrm{K} \) is mean energy per molecule of gas.
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