9
$\begingroup$

For an adiabatic process $pV^{\gamma}=k$ and after differentiating the equation we get that the slope of an adiabatic process is $$\frac{\mathrm dp}{\mathrm dV}=-{\gamma}\frac{\mathrm p}V$$

For an isothermal process $pV=k$ and after differentiating the equation we get that the slope of an isothermal process is $$\frac{\mathrm dp}{\mathrm dV}=-\frac pV$$

It says in my book that during compression adiabatic curve is above the isothermal curve and that during expansion the vice versa happens. My question is that since the slope of the adiabatic process is $\gamma$ times more than the slope of the isothermal process, shouldn't the curve of the former process always be above the latter.

$\endgroup$
0
3
$\begingroup$

The two cases you refer to seem to be for the reversible adiabatic and reversible isothermal change in an ideal gas.

In an adiabatic expansion as there is no heat loss or gain to or from the surroundings ($q=0$) the energy needed for the work of expansion against the external pressure must come from the internal energy of the gas. This results in a lowering of the temperature.

A curve of reversible adiabatic p vs. T change must therfore be below that for a reversible isothermal change. In addition as $\gamma \gt 1$, for example $5/3$ for an monoatomic gas, the adiabatic curve becomes even further below the isothermal one. The slope of the curves is not relevent in determining which is greater.

adiabatic p-v

$\endgroup$
1
  • 1
    $\begingroup$ Sorry but I believe the question was about the slope of a P-V curve (not p-T curve). (Context about me: My book also states the same results as the OP, and I too end up at the same doubt as he did.) $\endgroup$ Nov 3 '17 at 12:07
3
$\begingroup$

I think you have the answer in front of you already.

For an adiabatic process, $$PV^{\gamma}=k$$ and indeed, $$\frac{\mathrm{d}P}{\mathrm{d}V}=-{\gamma}\frac{P}{V}$$

For an isothermal process, $$PV=k$$ and again, $$\frac{\mathrm{d}P}{\mathrm{d}V}=-\frac{P}{V}$$

So the adiabatic curve is always steeper than an isothermal curve. (Note, $\gamma > 1$)

It says in my book that during compression adiabatic curve is above the isothermal curve and that during expansion the vice versa happens.

Okay, this is very easy to see. Just consider a toy problem.

Say, you have an arbitrary system that starts of at,

$$P_i = 2\ \text{and}\ V_i = 2$$

For the isothermal case, we can write $P$ as a function of $V$, $P(V) = \frac{4}{V}$ (here $k= P_i V_i = 4$)

Similarly, we can write $P(V) = \frac{2 \times 2^{5/3}}{V}$

I have used $\gamma = 5/3$. Now, say the gas expands do a final volume $V_f = 8$, you obtain a curve like the one shown below by plotting the functions we have just written down.

Now, say we start compressing the gas, starting at an initial volume and pressure, $V_i = 8, P_i = 0.5$ Isothermally, we trace the same curve back. Adiabatically, we obtain the green curve, since now $P(V) = \frac{0.5\ \times \ 8^{5/3}}{V}$

enter image description here

$\endgroup$
0
$\begingroup$

Your calculations are perfect.

It's just the starting point that matters... I believe that this diagram is self-explanatory:

enter image description here

(Sorry for the hand drawn diagram, I couldn't find the right graph online.)

$\endgroup$
0
$\begingroup$

An interpation of the implicit derivative by the marginal rate of substiution makes this clear

Marginal rate of substitution In economics, when the level set R(x, y) = 0 is an indifference curve for the quantities x and y consumed of two goods, the absolute value of the implicit derivative dy / dx is interpreted as the marginal rate of substitution of the two goods: how much more of y one must receive in order to be indifferent to a loss of one unit of x.- From wiki

Going back to our expression, $$PV^{\gamma} - C = 0$$This cast as an implicit function of the from $F(V,P)=0$, now $ \frac{dP}{dV}$ is how much more of pressure should increase to be indifferent to a loss of a unit volume (say on liter is the unit of volume).

In an adiabatic curve, we find the dP/dV is greater than that of isothermal case, this means we need more drop of pressure to account for a drop in unit volume.

Also check from 3:45 of this video

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.