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 statements 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)

(a) Work done = Area under $p-V$ graph
$A_1>A_2$
$\therefore W_{\text{given process}>W_{\text{isotermal process}}}$
(b) In the given process $p_V$ equation will be of a straight line with negative slope and positive intercept i.e.,
$p=-\alpha V+\beta$ (Here $\alpha$ and $\beta$ are positive constants)
$\Rightarrow p_V=-\alpha V^2+\beta V$
$\Rightarrow nRT=-\alpha V^2+\beta V$
$\Rightarrow T=\frac{1}{nR}(-{\alpha }^2+\beta V)$...(i)
This is an equation of parabola in $T$ and $V$
(d) $\frac{dT}{dV}=0-\beta -2\alpha V$
$\Rightarrow V=\frac{B}{2\alpha }$
Now, $\frac{d^{2}T}{d{V}^2=-2\alpha =-ve}$
ie, $T$ has some maximum value.
Now, $T\propto pV$ and $(pA{)}_A=(pV{)}_B$
$\Rightarrow T_A=T_B$
We conclude that temperatures are same at $A$ and $B$ and in between temperature has a maximum value.
Therefore, in going from $A$ to $B,T$ will first increase to a maximum value and then decrease.