A) \[{{n}_{1}}={{n}_{2}}\]
B) \[2{{n}_{1}}={{n}_{2}}\]
C) \[{{n}_{1}}=2{{n}_{2}}\]
D) \[2{{n}_{1}}=3{{n}_{2}}\]
Correct Answer: A
Solution :
\[{{f}_{av}}=\frac{total\text{ }number\text{ }of\text{ }degrees\text{ }of\text{ }freedom}{total\text{ }number\text{ }of\text{ }molecules}\] \[=\frac{{{n}_{1}}{{N}_{A}}{{f}_{1}}+{{n}_{2}}{{N}_{A}}{{f}_{2}}}{{{n}_{1}}{{N}_{A}}+{{n}_{2}}{{N}_{A}}}\] \[=\frac{{{n}_{1}}{{f}_{1}}+{{n}_{2}}{{f}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] \[=\frac{3{{n}_{1}}+5{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] Also \[\gamma =1+\frac{2}{{{f}_{av}}}=1.5\] Or \[{{f}_{av}}=4\] \[\therefore \] \[\frac{3{{n}_{1}}+5{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}=4\] Or \[{{n}_{1}}={{n}_{2}}\]
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