Question

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.
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 1

Correct Answer - D
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]`.