Question

An ideal gas is taken from state A (Pressure P, Volume V ) to the state B (Pressure P/2, Volume 2V) along a straight line path in PV diagram as shown in the adjacent figure.
Select the correct statement (s) among 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 the isotherm
• B. in the T-V diagram, the path AB become part of parabola
• C. in the P-T diagram, the path AB become part of hyperbola
• D. in going from A to B, the temperature T gas of the gas first increases to a maximum values 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 the isotherm
B in the T-V diagram, the path AB become part of parabola
D in going from A to B, the temperature T gas of the gas first increases to a maximum values then decreases

Work done $=$ Area under $P-V$ graph
and as isothermal process follow the hyperbolic path,
so $A_1$(area for the given process) > $A_2$(area for the isothermal curve)
$\therefore W_{\text{given process}}>W_{\text{isothermal process}}$
Thus (A) is correct.
In the given process $P-V$ equation will be a straight line with negative slope and positive intercept, i.e.,
$P=-aV+b$
Here, $a$ and $b$ are positive constants.
$P=-aV+b$
$\Rightarrow PV=-a{V}^2+bV$
$\Rightarrow nRT=-a{V}^2+bV[\because PV=nRT\left(\text{From ideal gas equation}\right)]$
$\Rightarrow T=\frac{1}{nR}(-a{V}^2+bV)$
The above equation represents a parabola.
Thus (B) is correct.
$T=\frac{1}{nR}(-a{V}^2+bV)$
$\frac{dT}{dV}=0$
$\Rightarrow \frac{1}{nR}(-2aV+b)=0$
$\Rightarrow -2aV+b=0$
$\Rightarrow V=\frac{b}{2a}$
Now,
$\frac{d^{2}T}{d{V}^2}=-2a=-ve$
$\therefore T$ has some maximum value.
Now, from ideal gas equation,
$T\propto PV$
$\because {\left(PV\right)}_A={\left(PV\right)}_B$
$\therefore T_A=T_B$
We conclude that temperatures are same at $A$ and $B$ but in between the temperature has a maximum value.
Hence in going from $A$ to $B$, the temperature will first increase to a maximum value and then decrease.