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                                    The volume of an ideal gas `(gamma 1.5 )` is changed adiabatically from 4.00 liters to 3.00 liters . Find the ratio of (a) the final pressure to the initial pressure and (b) the final temperature to the initial temperature.A.B.C.D. | 
                            
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Answer» `gamma=1.5` Since it is an adiabatic process, So, `PV^(gamma) =constant`. (a) `P_1V_1^(gamma) = P_2V_2^(gamma)` Given, `V_1 = 4L`, `V_2 = 3L`, `(P_2)/(P_1)=?` implies`P_2/P_1=((V_1)/(V_2))^gamma` `=(4/3)^(15)=1.5396=1.54` (b) `TV^(gamma-1) =Constant` `T_1V_1^(gamma-1) = T_2V_2^(gamma-1)` implies ` T_2/T_1=((V_1)/(V_2))^(gamma-1)=((4)/(3))^(0.5)=1.154`.  | 
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