Question

An ideal gas undergoes a cyclic process $abcda$ which is shown by a pressure - density curve. Identify the option with incorrect statement.

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

Answer 1

The correct option is C Internal energy of the gas at state $a$ is greater than that at state $c$

From the ideal gas equation,
$PM=\rho RT$
$\Rightarrow \frac{P}{\rho }=\frac{RT}{M}$
Here, $\frac{RT}{M}$ is the slope of $P-\rho$ curve.
As $R$ and $M$ are constant for a gas,
Slope $\left(m\right)\propto T$
From graph, $m_{cd}>m_{ab}$, thus, $T_c>T_a$
Also, the slope of the processes $ab$ and $cd$ are constant, thus, temperature is constant in these processes.
$\Rightarrow$ $ab$ and $cd$ are isothermal processes.
For a given gas, $\rho =\frac{M}{V}$.
For processes $bc$ and $da$, density is constant, thus, volume will be constant.
$\Rightarrow$ $bc$ and $da$ are isochoric processes.
Plotting the $P-V$ diagram, we get,


For option $\left(a\right)$,
In process $bc$, as there is no change in volume, the work done by gas is zero.
For option $\left(b\right)$,
In process $cd$, as volume is decreasing and pressure is increasing, thus it is a compression process. Therefore work done by gas is negative.
For option $\left(c\right)$,
As $T_c>T_a$, internal energy at $c$ is greater than at $a$.
For option $\left(d\right)$,
As the process is cyclic anticlockwise, thus, the work done by gas is negative.
Hence, option $\left(c\right)$ contains incorrect statement.

$\begin{array}{c}\text{Why this question?}\\ \text{It provides unique challenge to apply laws of thermodynamics and ideal gas equation.}\\ \text{Caution: For comparing work done and change in internal energy in a given process, always convert given curve to P - V diagram.}\end{array}$