Question

Two moles of an ideal gas changes from its initial state (16 atm, 6 L) to final state (4 atm, 15 L) in such a way that this change can be represented by a straight line on the P-V curve. The maximum temperature attained by the gas during the above change is:
$(\text{Take}R=\frac{1}{12}{\text{L atm K}}^{-1}{\text{mol}}^{-1})$

• A. 324 K
• B. 648 K
• C. 1296 K
• D. 972 K

Answer 1

The correct option is B 648 K

Equation of line:
$\frac{P-16}{16-4}=\frac{V-6}{6-15}$
Or $P-16=\frac{-12}{9}(V-6)=\frac{-4}{3}V+8$
Or $P=24-\frac{4}{3}V$
$\therefore PV=nRT=(24-\frac{4}{3}V)V=24V\frac{-4V^2}{3}$
$@\text{maximum}T,\frac{dT}{dV}=0\Rightarrow 24\frac{-8V}{3}=0$
$\therefore V=\frac{24\times 3}{8}=9L(P=24-\frac{4}{3}\times 9=12atm)$
$\therefore T_{max}=\frac{PV}{nR}$
$=\frac{12\times 9}{2\times \frac{1}{12}}=648K$