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.

Options

A. `P_(0) V_(0)[In 2 + (1)/(2)]`
B. `P_(0) V_(0)[In 2 + (3)/(2)]`
C. `P_(0) V_(0)[In 2 + 2]`
D. `P_(0) V_(0)[In 2 - (1)/(2)]`

Answer

D

Explanation

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_0)/(V_0) = nRT_(0)1n2=p_(0)V_(0)1n2`
Also from the ideal gas equation,
`Pa V_(a) = p_(b) V_(b)` or `p_b = (p_a V_a)/(V_b) = (p_0 V_0)/(2V_0) = (p_0)/(2)`
in the step `bc` the pressure remain constant . Hence the work done is `W_(2) = (p_0)/ (V_0-2V_0) = -(P_0V_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 2-1/2]`.