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

$$\begin{align} T_\mathrm f&=360\ \mathrm K\\ T_\mathrm i&=576\ \mathrm K \end{align}$$ I tried finding a function in $T$ which I could maximise, but in vain.

  • $\begingroup$ The function to maximize is the ideal gas law. $\endgroup$ – a-cyclohexane-molecule Aug 8 '16 at 13:50
  • $\begingroup$ I understand that, but I am unable to apply the ideal gas law FOR the change that occurs. $\endgroup$ – Akshar Gandhi Aug 8 '16 at 13:51
  • $\begingroup$ I don't understand; please specify. Why are you unable to apply the ideal gas law? $\endgroup$ – a-cyclohexane-molecule Aug 8 '16 at 14:06
  • $\begingroup$ I am actually unable to identify where I should apply the equation. $\endgroup$ – Akshar Gandhi Aug 8 '16 at 14:07

T is maximum when PV is maximum. So get the equation for the straight line through the two end points, expressed in the form $P=aV+b$. Then write $PV=aV^2+bV$. Find the maximum of the function $aV^2+bV$. It is at $V=-b/(2a)$. So the maximum of PV is $$(PV)_{max}=-\frac{b^2}{4a}$$ This is equal to $nRT_{max}$

From fitting the straight line to the end points, the slope obtained is a = -4/3, and the intercept is b = 24. So, from the equation above, the maximum value of PV is $$(PV)_{max}=108\ liter-atm$$


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.

  • $\begingroup$ I still cannot understand your method. Would you please post a complete solution with answer? Thanks 👍 $\endgroup$ – Akshar Gandhi Aug 8 '16 at 14:37
  • 1
    $\begingroup$ Very well. I will try my utmost to understand. Thanks for all your help! $\endgroup$ – Akshar Gandhi Aug 8 '16 at 15:22

above question can be solved by the method given in the picture



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