A) 1.4
B) 1.55
C) 1.62
D) 1.67
Correct Answer: B
Solution :
For monoatomic gas \[{{C}_{V}}=\frac{3}{2}R\], From Mayors formula \[{{C}_{p}}-{{C}_{V}}=R\] \[\Rightarrow \] \[{{C}_{p}}=R+\frac{3}{2}R=\frac{5}{2}R\] For diatomic gas \[{{C}_{V}}=\frac{5}{2}R\], and \[{{C}_{p}}=R+{{C}_{V}}=R+\frac{5}{2}R=\frac{7}{2}R\] For the mixture \[Q=n{{C}_{V}}\,\,\Delta t\], we have \[3{{C}_{V}}\,\Delta t=2\times \frac{3}{2}R\times \Delta t+1\times \frac{5}{2}R\times \Delta t\] \[\Rightarrow \] \[{{C}_{V}}=\frac{11}{6}R\] The value of \[{{C}_{p}}\]for mixture is \[2\times \frac{5}{2}R\times \Delta t+1\times \frac{7}{2}R\times \Delta t=3{{C}_{p}}\Delta t\] \[\Rightarrow \] \[{{C}_{p}}=\frac{17R}{6}\] \[\therefore \] \[\gamma =\frac{{{C}_{p}}}{{{C}_{V}}}=\frac{17R}{6}\times \frac{6}{11R}=1.55\]
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