A) 250 K
B) 500 K
C) 1000 K
D) 1500 K
Correct Answer: A
Solution :
[a]: Degree of freedom associated with rotational motion of diatomic gas molecules = 2 |
Average rotational energy per molecule |
\[=l{{\omega }^{2}}_{rms}=2\times \frac{1}{2}kT=kT\] |
Hence initial temperature \[{{T}_{1}}\] of diatomic gas |
\[=\frac{1}{2}l\omega _{0}^{2}=k{{T}_{1}}\] which give \[{{T}_{1}}=250\,K\] |
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