A) \[{{n}_{1}}={{n}_{2}}\]
B) \[2{{n}_{1}}={{n}_{2}}\]
C) \[3{{n}_{1}}=2{{n}_{2}}\]
D) \[2{{n}_{1}}=3{{n}_{2}}\]
Correct Answer: A
Solution :
From the formula \[{{f}_{av}}=\frac{Total\text{ }number\text{ }of\text{ }degree\text{ }of\text{ }freedom}{Total\text{ }number\text{ }of\text{ }molecules}\] \[=\frac{{{n}_{1}}{{N}_{A}}{{f}_{1}}+{{n}_{2}}{{N}_{a}}{{f}_{2}}}{{{n}_{1}}{{N}_{A}}+{{n}_{2}}{{N}_{A}}}\] \[\Rightarrow \] \[{{f}_{av}}=\frac{{{n}_{1}}{{f}_{1}}+{{n}_{2}}{{f}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] From the relation \[\gamma =1+\frac{2}{{{f}_{av}}}\] Give : \[\gamma =1.5\] So, \[1.5=1+\frac{2}{{{f}_{av}}}\] or \[{{f}_{av}}=4\] So, \[\frac{{{S}_{n1}}+{{S}_{n2}}}{{{n}_{1}}+{{n}_{2}}}=4\] or \[{{n}_{1}}={{n}_{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
