Question

$n$ moles of an ideal gas undergoes a process $A \to B$ as shown in the figure. The maximum temperature of the gas during the process will be:

A. $\dfrac{{9{P_0 }{V_0 }}}{{4nR}}$
B. $\dfrac{{9{P_0 }{V_0 }}}{{2nR}}$
C. $\dfrac{{9{P_0 }{V_0 }}}{{nR}}$

Answer 1

Here we shall obtain the equation of the given straight line in the graph using the two-point formula of a straight line equation. We shall substitute this equation in the ideal gas equation to eliminate either the pressure $P$, or the volume $V$ to obtain an equation having temperature and another variable (here volume). To maximize the temperature we shall differentiate the equation.

Formula used: $PV = nRT$

Complete step by step answer:
From the graph given above, we can find the equation of the straight line using a two-point formula. The equation of the line is
$\dfrac{{P - 2{P_0 }}}{{V - {V_0 }}} = \dfrac{{2{P_0 } - {P_0 }}}{{{V_0 } - 2{V_0 }}}$.
$\Rightarrow \dfrac{{P - 2{P_0 }}}{{V - {V_0 }}} = \dfrac{{{P_0 }}}{{ - {V_0 }}}$
Upon rearranging the equation we get,
$\Rightarrow P = - \dfrac{{{P_0 }}}{{{V_0 }}}(V - {V_0 }) + 2{P_0 }$
$\Rightarrow P = - V\dfrac{{{P_0 }}}{{{V_0 }}} + 3{P_0 }$
Now from the ideal gas equation, we get
$PV = nRT$
where $P$ is the pressure of the gas, $V$ is the volume of the gas, $n$ is the number of moles of the ideal gas, $R$ is the real gas constant, and $T$ is the absolute temperature.
Substituting the equation of pressure $P$, as obtained previously into this ideal gas equation, we have,
$( - V\dfrac{{{P_0 }}}{{{V_0 }}} + 3{P_0 })V = nRT$.
Thus we have obtained an equation of the temperature $T$ in terms of a single variable $V$, the volume of the ideal gas. Thus to maximize the temperature, over the entire process $A \to B$, we have to differentiate the above equation with respect to the volume $V$.
Thus, from $( - V\dfrac{{{P_0 }}}{{{V_0 }}} + 3{P_0 })V = nRT$, we have
$\dfrac{d}{{dV}}( - V\dfrac{{{P_0 }}}{{{V_0 }}} + 3{P_0 })V = nR\dfrac{{dT}}{{dV}}$
$\Rightarrow ( - V\dfrac{{{P_0 }}}{{{V_0 }}} + 3{P_0 }) + V( - \dfrac{{{P_0 }}}{{{V_0 }}}) = nR\dfrac{{dT}}{{dV}}$
Upon simplifying we get,
$\Rightarrow nR\dfrac{{dT}}{{dV}} = - 2V\dfrac{{{P_0 }}}{{{V_0 }}} + 3{P_0 }$
Since we are maximizing the temperature $T$ over the volume $V$ within the process, we will have $\dfrac{{dT}}{{dV}}$ as $0$, since the temperature will reach its peak value, which we need to obtain.
Thus substituting $\dfrac{{dT}}{{dV}} = 0$, in the above equation, we have,
$\Rightarrow nR(0) = - 2V\dfrac{{{P_0 }}}{{{V_0 }}} + 3{P_0 }$
$\Rightarrow 0 = - 2V\dfrac{{{P_0 }}}{{{V_0 }}} + 3{P_0 }$
On simplifying and rearranging, we get,
$\Rightarrow 2V\dfrac{{{P_0 }}}{{{V_0 }}} = 3{P_0 }$
$\Rightarrow V = \dfrac{3}{2}{V_0 }$
We also check that this value is within the limits of the process, i.e. within $A$ and $B$.
Substituting this value of $V$ in the equation of the straight line, we have the pressure value as,
$\Rightarrow P = - (\dfrac{3}{2}{V_0 })\dfrac{{{P_0 }}}{{{V_0 }}} + 3{P_0 }$
$\Rightarrow P = \dfrac{3}{2}{P_0 }$
We also check that this value is within the limits of the process, i.e. within $A$ and $B$.
Thus now substituting both these values of pressure and temperature in the ideal gas equation, we have
$PV = nRT$
$\Rightarrow (\dfrac{3}{2}{P_0 })(\dfrac{3}{2}{V_0 }) = nR{T_{\max }}$
where ${T_{\max }}$ is the maximum temperature which is attained at the substituted values of pressure and volume.
$\Rightarrow {T_{\max }} = (\dfrac{3}{2}{P_0 })(\dfrac{3}{2}{V_0 })\dfrac{1}{{nr}}$
$\Rightarrow {T_{\max }} = \dfrac{{9{P_0 }{V_0 }}}{{4nR}}$
Thus, the correct answer is option (A).

Note: On differentiating the equation of temperature and volume, we obtained an equation where we have substituted $\dfrac{{dT}}{{dV}} = 0$. This happens at both the maxima and minima of the temperature. To check whether the point is really a maxima, we have to obtain $\dfrac{{{d^2}T}}{{d{V^2}}}$ and which we get equal to $- \dfrac{{2V}}{{nR}}\dfrac{{{P_0 }}}{{{V_0 }}}$. Since this value is negative, thus the concerned temperature is the maximum.