A sample of an ideal gas has pressure `p_(0)`, volume `V_(0)` and tempreture `T_(0)`. It is isothermally expanded to twice its oringinal 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 tempreture.(a) show the process in a V-T diagram (b) calculate the heat absorbed in the process.

Explanation

(a) The V-T diagram for the process is shown in . The initial state is represent by the point a. in the first step, it is isothermally expanded to a volume `2V_((0))`. This is shown by ab. Then the pressure is kept constant and the gas is compressed to the volume to `V_((0))`. form the ideal gas equetion, `V//T` is constant att constant pressure. Hence, the oorigin. At point c, the volume is `V_((0))`. in the final step, the gas is heated at constant volume to a tempreture `T_((0))`. this is shown by ca. the final state is the same as the inital state.(b) the process is cycle so that the change in internal energy is zero.the heat supplied is, therefore, equel to the work done by the gas. the work done during ab is
`W_(1)=nRT_(0)In(2V_(0))/(V_(0))=nRT_((0))In 2=p_((0))V_((0))In2`.
Also from the ideal gas equetion,
`p_(a)V_(a)=p_(b)V_(b)`
`p_(a)V_(a)=(p_bV_b)/V_(b)=(p_(0)V_(0))/(2V_(0))=p_(0)/(2)`
in the step bc, the pressusre remains constant hence, the work done is,
`W_(2)=p_((0))/(2)(V_((0))-2V_((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 cycle process is
`W=W_(1)+W_(2)`
`=p_((0))V_((0))[In 2-0.5]`
`=0.193 p_((0))V((0)).
hence, the heat suplied to the gas is `0.193p_((0))V_((0))`.