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 in the pressure-volume diagram. Select the correct statements from the following.

• A. The work done by the gas in the process $A$ to $B$ exceeds the work done that would be done by it if the system were taken from $A$ to $B$ along an isotherm.
• B. In the temperature-volume diagram, the path $AB$ becomes a part of a parabola.
• C. In the pressure-temperature 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)

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

The given process follows a linear P-V relation given by: $\frac{p-P}{v-V}=\frac{P-P/2}{V-2V}=-P/2V$
$\Rightarrow p-P=\frac{-P}{2V}(v-V)\Rightarrow p=-\frac{Pv}{2V}+3P/2$ .......(i)

Here work done $\Delta W$=area under P-V diagram =$3PV/4=0.75PV$
For isotherm, pv=PV $\Rightarrow$ work done $\Delta W$=$PV\ln 2=0.693PV$

Hence statement a is correct
For t-v diagram, replace p in (i) by $\frac{nRt}{v}$
Hence, $t=-\frac{P{v}^2}{2nRV}+\frac{3Pv}{2nR}$

The above eqn. has t equalling a quadratic in v.
Hence, t-v diagram is a parabola $\Rightarrow$ statement b is correct.

Clearly the above parabola is concave downwards , hence has a maxima.
The t has same initial and final value implying the maxima occurs in-between.

Hence statement d is correct.