Question

One mole of an ideal gas is carried through a reversible cyclic process as shown in the figure. The maximum temperature attained by the gas during the cycle is (in K):

• A. $\frac{7}{6R}$
• B. $\frac{12}{49R}$
• C. $\frac{49}{12R}$
• D. None of these

Answer 1

The correct option is C $\frac{49}{12R}$

Highest T will be attained between 'B' and 'C'.
Line B to C $:\frac{P-4}{4-1}=\frac{V-1}{1-2}$
$\Rightarrow P-4=-3V+3\Rightarrow P=7-3V$
$\therefore PV(=nRT)=(7-3V)V=7V-3V^2$
Maxima when $\frac{d\left(PV\right)}{dV}=zero.$
$\Rightarrow 7-6V=0orV=\frac{7}{6}\therefore P=7-\frac{7}{2}=\frac{7}{2}$
$\therefore T_{max}=\frac{PV}{nR}=\frac{\frac{7}{2}\times \frac{7}{6}}{R}=\frac{49}{12R}K$