Question

An ideal gas undergoes a cyclic process $\text{abcda}$ which is shown by the pressure - density curve.

• A. Work done by the gas in the process $bc$ is zero.
• B. Work done by the gas in the process $cd$ is negative.
• C. Internal energy of the gas in state $a$ is greater than in state $c$.
• D. Net work done by the gas in the cycle is negative.

Answer 1

Answer: (A), (B), (D)

The correct options are
A Work done by the gas in the process $bc$ is zero.
B Work done by the gas in the process $cd$ is negative.
D Net work done by the gas in the cycle is negative.

Ideal gas equation can be written in terms of density as
$P=\frac{\rho }{M}RT\Rightarrow \frac{P}{\rho }=\frac{R}{M}T$
[Using $\rho =\frac{m}{V}$]
From the given graph, we can infer that for process $cd$ and $ab$, slope of the graph remains constant.
$\therefore$ Temperature remains constant.
Hence, $cd$ and $ab$ are isothermal processes.
As $\rho \propto \frac{1}{V}$, we can also infer from the graph that $bc$ and $da$ are isochoric processes.


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 $a\to b\to c\to d\to a$, the net work done by the gas is negative,

Also, from $d\to a$, temperature decreases since $P\propto T$. So, internal energy decreases from $d\to a$. 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.