Find the maximum temperature attained by the gas during the change The main question is:


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


My approach :
I plotted the initial and final conditions on a $P{-}V$ curve using a Cartesian coordinate system. Joining them gives me the temperature transition.

I also made two isotherms (hyperbolic curves) on the same graph, passing through the points a previously plotted. The values of initial and final temperatures I calculated are 

$$\begin{align}
T_\mathrm f&amp;=360\ \mathrm K\\
T_\mathrm i&amp;=576\ \mathrm K
\end{align}$$
I tried finding a function in $T$ which I could maximise, but in vain.
The ideal gas law $pV = nRT$ allows us to relate one of $p$, $V$, and $T$ to the two other variables, and is the main equation to work with here. 

Rephrasing the original question to make it more suggestive, we are interested in maximizing $T(p,V)$ subject to the constraint $p(V)$, a linear function in $V$ readily found from the conditions specified in the problem. Substituting $p(V)$ for $p$ to get $T = T(p(V),V) = T(V)$, we have reduced the problem to one amenable to simple calculus techniques, and the answer follows from taking a derivative.



Remark. In general, extremization problems are not so easy, and we will need to use the more powerful tool of Lagrange multipliers.