Question

$n$ moles of an ideal gas undergoes a process $A\to B$ as shown in the diagram. The maximum temperature of the gas during the process is

• A. $\frac{3{\text{P}}_0{\text{V}}_0}{2\text{nR}}$
• B. $\frac{9{\text{P}}_0{\text{V}}_0}{4\text{nR}}$
• C. $\frac{9{\text{P}}_0{\text{V}}_0}{2\text{nR}}$
• D. $\frac{9{\text{P}}_0{\text{V}}_0}{\text{nR}}$

Answer 1

Answer: (B)

Since the P-V graph of the process is a straight line and two points (V₀, 2P₀) and (2V₀, P₀) are known, its equation will be
$\left(\text{P}-{\left(\text{P}\right)}_0\right)=\frac{\left(2{\left(\text{P}\right)}_0-{\left(\text{P}\right)}_0\right)}{\left({\left(\text{V}\right)}_0-2{\left(\text{V}\right)}_0\right)}\left(\text{V}-2{\left(\text{V}\right)}_0\right)=\frac{{\left(\text{P}\right)}_0}{{\left(\text{V}\right)}_0}\left(2{\left(\text{V}\right)}_0-\text{V}\right)$
$\therefore \text{P}=3{\text{P}}_0-\frac{{\text{P}}_0\text{V}}{{\text{V}}_0}$
According to the equation for an ideal gas,
$\text{T}=\frac{\text{pV}}{\text{nR}}$
$=\left(3{\left(\text{P}\right)}_0-\frac{{\left(\text{P}\right)}_0\text{V}}{{\left(\text{V}\right)}_0}\right)\frac{\text{V}}{\text{nR}}$
$=\frac{3{\text{P}}_0{\text{V}}_0\text{V}-{\text{P}}_0{\text{V}}^2}{\text{nR}{\text{V}}_0}$ ...(i)
For $T$ to maximum, $\frac{\text{dT}}{\text{dV}}=0$
$3{\text{P}}_0{\text{V}}_0-2{\text{P}}_0\text{V}=0$
or $\text{V}=\frac{3{\text{V}}_0}{2}$ ...(ii)
Putting this value in Eq. (i), we get
${\left(\text{T}\right)}_{\text{max}}=\frac{3{\left(\text{P}\right)}_0{\left(\text{V}\right)}_0\left(\frac{3{\left(\text{V}\right)}_0}{2}\right)-{\left(\text{P}\right)}_0\left(\frac{9}{4}{\text{V}}_0^2\right)}{\text{nR}{\left(\text{V}\right)}_0}=\frac{9{\left(\text{P}\right)}_0{\left(\text{V}\right)}_0}{4\text{nR}}$