Question

An ideal gas is taken from the state $A$ (pressure $P$, volume $V$) to the state $B$ (pressure $P/2$, volume $2V$) along a straight line path 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 a 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 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
D In going from $A$ to $B$, the temperature $T$ of the gas first increases to a maximum value and then decreases

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 the isotherm. This is because the work done is the area under the $P-V$ indicator diagram. As shown, the area under the graph in the first diagram will be more than that in the second diagram. When we extrapolate the graph shown in Fig. let $P_0$ be the intercept on the $P-axis$ and $V_0$ be the intercept on the $V-axis$. The equation of the line $AB$ can be written as
$P=-\frac{P_0}{V_0}V+P_0[\because y=mx+c]...\left(i\right)$
To find a relationship between $P$ and $T$, we use
$PV=RT\Rightarrow V=\frac{RT}{P}....\left(ii\right)$
From Eqs. (i) and (ii),
$P=-\frac{P_0}{V_0}\times \frac{RT}{P}+P_0$
$\Rightarrow P^2{V}_0-P{P}_0{V}_0=-P_{0}RT...\left(iii\right)$
Relation between $P$ and $T$ is the equation of a parabola.
Also $PV=RT$
$\therefore P=\frac{RT}{V}....\left(iii\right)$
From Eqs. (i) and (ii),
$\frac{RT}{V}=-\frac{P_0}{V_0}V+P_0$
$\Rightarrow RT=-\frac{P_0}{V_0}{V}^2+P_{0}V....\left(iv\right)$
The above equation is of a parabola (between $T$ and $V$)
$T=-\frac{P_0}{V_{0}R}{V}^2+\frac{P_0}{R}V$
Differentiating the above equation w.r.t. $V$ we get
$\frac{dT}{dV}=-\frac{P_0}{V_{0}R}\times 2V+\frac{P_0}{R}$
when $\frac{dT}{dV}=0$,
then $\frac{P_0}{V_{0}R}\times 2V\frac{P_0}{R}\Rightarrow V=\frac{V_0}{2}$
Also $\frac{d^{2}T}{d^{2}V}=\frac{-2P_0}{V_{0}R}=-ve$
$\Rightarrow V=V_0/2$ is the value of maxima of temperature
Also $P_A{V}_A=P_B{V}_B\Rightarrow T_A=T_B$ (From Boyle's law)
$\Rightarrow$ In going from $A$ to $B$, the temperature of the gas first increases to a maximum (at $V=V_0/2)$ and the decreases and reaches back to the same value.