A sample of an ideal gas has pressure `p_(0)`, volume `V_(0)` and tempreture `T_(0)`. It is isothermally expanded to twice its original volume.It is then compressed at constant pressure to have the original volume `V_(0)`. Finally, the gas is heated at constant volume to get the original temperature.
(a) Show the process in a V-T diagram
(b) Calculate the heat absorbed in the process.

Explanation

The V-T diagram for the process is shown in figure. The initial state is represented by the POINT a. In the first step, it is isothermally expanded to a volume `2V_(0)`. This shown by ab. Then the pressure is kept constant and the gas is compressed to the volume V0. From the ideal gas equation, `V//T` is constant at constant pressure. Hence, the process is shown by a line bc which is `V_(0)`. In the final step, the gas is heated at constant volume to a temperature `T_(0)`. This is shown by ca. The final state is the same as the initial state.
b. The process is cyclic so that the change in internal energy is zero. The heat supplied is, therefore, equal to the work done by the gas. The work done during ab is
`W_(1) = nRT_(0) 1n (2V)/(V_(0)) = nRT_(0) 1n 2 = p_(0) V_(0) 1n 2`
Also from the ideal gas equation.
`P_(a) V_(a) = P_(b) V_(b)` or `P_(b) = (P_(a) V_(a))/(V_(b)) = (P_(0) V_(0))/(2 V_(0)) = (P_(0))/(2)`
In the step bc, the pressure remains constant. Hence the
work done is, `W_(2) = (P_(0))/(2) (V_(0) - 2 V_(0)) = - (P_(0) V_(0))/(2)`
In the step ca, the volume remains constant and so the work done is zero. The net work done by the gas in the cyclic process is
`W = W_(1) + W_(2)`
`= P_(0) V_(0) [1n 2 - 0.5]`
`0.193 P_(0) V_(0)`.