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Monoatomic gas A having 5 mole is mixed with diatomic gas B having 1 mole in container of volume `V_(0)`. Now the volume of mixture is compressed to `(V_(0))/(4)` by adiabatic process. Initial pressure and temperature of ags mixture is `P_(0)` and `T_(0)`. [given `2^(3.2)=9.2`] Choose correct option :A. `._(gamma mix)=1.6`B. Final pressure is between `9P_(0)` and `10P_(0)`C. `|W.D|=13RT_(0)`D. none of these

Answer» `._(gamma mix)=(n_(1)C_(P1)+n_(2)C_(P2))/(n_(1)C_(V1)+n_(2)C_(V2))=(8)/(5)`
`W = (P_(1)V_(1)-P_(2)V_(2))/(gamma-1)`
`P_(0)V_(0)^(8//5)=P_(2)((V_(0))/(4))^(8//5)`
`P_(2)=9.2P_(0)`
`W = (P_(0)V_(0)-9.2P_(0)(V_(0))/(4))/(3//5)= - 13RT_(0)`


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