The initial volume of an ideal gas is `V_(1)` and the initial pressure is `P_(1)=6` atmosphere. It expands isothermally to a volume `V_(2)` and pressure `P_(2)=3 atm`, and then adiabatically to a volume `V_(3)` and pressure `P_(3)=2 atm`. Draw a rough sketch of these changes. Calculate the value of `V_(2)//V_(1) and V_(3)//V_(2)`.
Take `gamma= 1.4`.

Explanation

In (figure), initial state of gas is represented by A `(V_(1), P_(1))`. AB represents isothermal change to B `(V_(2),P_(2))`
BC represents adiabatic change from B`(V_(2), P_(2))` to `C(V_(3),P_(3))`.
As change from A to B is isothermal
`:. P_(1)V_(1)= P_(2)V_(2) or (V_(2))/(V_(1))= (P_(1))/(P_(2))= 6/3=2`
As change from B to C is adiabatic,
`:. P_(2)V_(2)^(gamma)=P_(3)V_(3)^(gamma) or ((V_(3))/(V_(2)))^(gamma)= (P_(2))/P_(3)`
or `(V_(3))/(V_(2))=(P_(2)/(P_(3)))^(1//gamma)=(3/2)^(1//gamma)= (1.5)^(1//gamma)= (1.5)^(1//1.4)=(1.5)^(5//7)=1.33`