Question

One mole of monoatomic ideal gas undergoes a process $AB{C}^{\prime }$ as shown in figure. The maximum temperature of the gas during the process $AB{C}^{\prime }$ is in the form $x\left(\frac{PV}{nR}\right)$. Find the value of $x$.

Answer 1

Temperature will be maximum when product of $P$ and $V$ is maximum. From the graph we can observe that, it will be maximum in the process $B{C}^{\prime }$.

The process $B{C}^{\prime }$ can be represented by straight line, $y=mx+c$
So, $P=mV+c$
Putting coordinates of $B$ & point $C^{\prime }$:
$3P=2mV+c$ ----(i)
$P=6mV+c$ ----(ii)
On substracting we get,
$2P=-4mV$
$\Rightarrow m=\frac{-P}{2V}$

From Equation (ii):
$P=6\times \left(\frac{-P}{2V}\right)V+c$
$\therefore c=4P$

Hence we can write the equation $y=mx+c$ in following form:
or, $y=\left(\frac{-P}{2V}\right)x+4P-----\left(iii\right)$
Here $y$ is pressure of gas and $x$ is the volume of gas ($xy=nRT$)

Multiplying with $x$ on both sides,
$xy=\frac{-P{x}^2}{2V}+4Px$
$\Rightarrow nRT=\frac{-P{x}^2}{2V}+4px-----\left(iv\right)$

For maximum temperature $\frac{dT}{dx}=0$
Differentially both sides w.r.t $x$:
$\frac{nRdT}{dx}=\frac{-2Px}{2V}+4P$
or, $\frac{dT}{dx}=\frac{\frac{-2Px}{2V}+4P}{nR}$

Applying condition of maximum temperature we get,
$\frac{-2Px}{2V}+4P=0$
$\therefore x=4V$

Substittuing in equation (iv):
$nR{T}_{max}=\frac{-P}{2V}(4V{)}^2+4P(4V)$
$\Rightarrow nR{T}_{max}=\frac{-16P{V}^2}{2V}+16PV=8PV$
$\Rightarrow T_{max}=\frac{8PV}{nR}$
Comparing with $T_{max}=x\left(\frac{PV}{nR}\right)$, we will get, $x=8$

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: In this problem focus on obtaining the releation}\\ \text{between P, V and T such that we can apply the condition for}\\ \text{obtaining maximum temperature.}\end{array}$