A) 3/5
B) 5/3
C) 1
D) none of these
Correct Answer: C
Solution :
| [c]: Average number of degree of freedom per molecule |
| \[f=\frac{\text{Total number of degree of freedom}}{\text{Total number of molecule}}\] |
| \[=\frac{{{n}_{1}}{{N}_{A}}{{f}_{1}}+{{n}_{2}}{{N}_{A}}{{f}_{2}}}{{{n}_{1}}{{N}_{A}}+{{n}_{2}}{{N}_{A}}}=\frac{{{n}_{1}}{{f}_{1}}+{{n}_{2}}{{f}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] |
| Here, \[{{f}_{1}}=3,\,{{f}_{2}}=5\] and \[\gamma =1+1/f\] |
| Also, \[\gamma =\frac{{{C}_{p}}}{{{C}_{v}}}=\frac{3R}{2R}\,\,=1.5=1+1\,/f\,\Rightarrow \,\,f=4\] |
| Hence, \[\frac{3{{n}_{1}}+5{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}=4\] which gives \[{{n}_{1}}/{{n}_{2}}=1\] |
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