Question

An ideal gas is taken from state $A$ (pressure $P,$ volume $V$) to state $B$ (pressure $\frac{P}{2}$, volume $2V$) along a straight line path as shown in the $P-V$ diagram.



Select the correct statement(s) from the following.

• A. The work done by the gas in the process $A$ to $B$ exceeds the work that would be done by it, if the system were taken from $A$ to $B$ along an isotherm.
• B. In the $T-V$ diagram, the path $AB$ becomes a part of a parabola,
• C. In the $P-T$ diagram, the path $AB$ becomes a part of hyperbola.
• D. In going from $A$ to $B$, the temperature $T$ of the gas first increases to a maximum value and then decreases.

Answer 1

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

In going from $A$ to $B$, the temperature $T$ of the gas first increases to a maximum value and then decreases.

Figure below shows the straight line path along with the corresponding isothermal path. Since the work done by the gas is equal to area under the curve (such as shown in the figure by the shaded portion for the isothermal path), it is obvious that the gas does more work along the straight line path as compared with that of isothermal path.


As the volume is increased from $V$ to $2V$, the difference of pressure between the straight line path and isothermal path initially increases and then decreases after attaining a maximum value. The same trend is observed in the case of temperature
$P\propto T,(\because$ $V$ is constant$)$
Now, the slope of straight line path is
$m=\frac{P-\frac{P}{2}}{V-2V}=-\frac{P}{2V}$
$P=-2Vm$
Putting this in the ideal gas equation,
$PV=nRT$
$[-2Vm]V=nRT$
$V^2=-\frac{nR}{2m}T$
$V^2=kT$
Which is the equation of a parabola.
Similarly, eliminating V from idea gas equation, we get
$P[-\frac{P}{2m}]=nRT$
$P^2=\text{(constant})T$
Which is again an equation of a parabola.

Why this question ? Key Concept: Work done is area under PV curve. Importance in JEE: General formula for parabola, hyperbola, ellipse, rectangular hyperbola are important for JEE.