A) 1.40
B) 1.50
C) 1.53
D) 3.07
Correct Answer: B
Solution :
Let,\[{{U}_{1}}\]be the internal energy of the monoatomic gas and the internal energy of the diatomic gas. Now, the internal energy of the mixture \[={{U}_{mix}}={{U}_{1}}+{{U}_{2}}.\] or,\[({{n}_{1}}+{{n}_{2}}){{C}_{{{v}_{mix}}}}={{n}_{1}}{{C}_{{{v}_{1}}}}T+{{n}_{2}}{{C}_{{{v}_{2}}}}T\] \[\Rightarrow \]\[{{C}_{{{v}_{mix}}}}=\frac{{{n}_{1}}{{C}_{{{v}_{1}}}}+{{n}_{2}}{{C}_{{{v}_{2}}}}}{{{n}_{1}}+{{n}_{2}}}=\frac{1\times \frac{3R}{2}+1\times \frac{5R}{2}}{1+1}=2R\] \[\therefore \] \[\frac{R}{{{\gamma }_{mix}}-1}=2R\] \[\Rightarrow \]\[{{\gamma }_{mix}}=\frac{3}{2}=1.5\] Hence, the correction option is [b].
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