The correct option is
D Net work done by the gas in the cycle is negative.
Ideal gas equation can be written in terms of density as
P=ρMRT⇒Pρ=RMT
[Using
ρ=mV]
From the given graph, we can infer that for process
cd and
ab, slope of the graph remains constant.
∴ Temperature remains constant.
Hence,
cd and
ab are isothermal processes.
As
ρ∝1V, we can also infer from the graph that
bc and
da are isochoric processes.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1004321/original_9s.png)
From the equivalent
P−V diagram,
Work done by the gas during processes
bc and
da is zero.
Work done by the gas during
cd is negative and the work done during
ab is positive.
Area of graph under
cd is greater than area of graph under
ab
So, we can conclude that in the cyclic process
a→b→c→d→a, the net work done by the gas is negative,
Also, from
d→a, temperature decreases since
P∝T. So, internal energy decreases from
d→a. But, internal energy at
c and
d are the same [isothermal]. Thus, internal energy at
a is lesser than at
c. Option (c) is wrong.
Thus, options (a) (b) and (d) are correct answers.