Question

The given figure shows an isochoric process, an isothermal process, an adiabatic process and two isobaric processes (one for monoatomic gas and other for diatomic) on a work done $\left(W\right)$ versus heat supplied$\left(Q\right)$ curve. The initial states of both gases are the same and the scales for the two axes are same.



Which of the following statements is incorrect?

• A. Straight line $1$ corresponds to an isochoric process.
• B. Straight line $2$ corresponds to an isobaric process for diatomic gas.
• C. Straight line $4$ corresponds to an isothermal process.
• D. Straight line $1$ corresponds to an isothermal process.

Answer 1

The correct option is D Straight line $1$ corresponds to an isothermal process.


From the first law of thermodynamics
$Q=W+\Delta U$
For an isobaric process
$W=P\Delta V=nR\Delta T$
$Q=n{C}_P\Delta T=n\left(\frac{f+2}{2}\right)R\Delta T$
$\Rightarrow \frac{W}{Q}=\frac{2}{f+2}$
where $f$ is the number of degrees of freedom, $\left(\frac{W}{Q}\right)$ represents the slope of $W-Q$ graph.
$\because$For monatomic gas $f=3$ and for diatomic gas $f=5$.
$\Rightarrow \frac{W}{Q}=\frac{2}{5}$ (for monatomic gas)
$\&\frac{W}{Q}=\frac{2}{7}$ (for diatomic gas)
Hence, for both the gases, isobaric curve will be straight line passing through origin and slope represented by value of their $\left(\frac{W}{Q}\right)$.

For an isochoric process,
$W=0,\frac{W}{Q}=0$
The straight line $1$ corresponds to an isochoric process because slope of $W-Q$ is zero.

For adiabatic process $Q=0$, hence line $5$ represents adiabatic process.

For isothermal process, $\Delta U=0,\text{hence}W=Q$
$\text{slope}=\tan \theta =\frac{W}{Q}=1$
$\therefore$ The isothermal process will have a straight line with highest slope compared to other processes.
$\Rightarrow$Straight line $4$ corresponds to isothermal process.
$\therefore$ Option $\left(d\right)$ is incorrect.