## Isothermal vs Adiabatic expansion

Evidently the adiabatic curve is steeper than the isothermal curve.

How can I prove this mathematically using the following facts:

$P_{1}V_{1} = P_{2}V_{2}$ for a reversible isothermal process
$P_{1}V^{γ}_{1} = P_{2}V^{γ}_{2}$ and $γ > 1$ for a reversible adiabatic process.

It's really a math problem but I don't know how to begin.

BiP

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 Quote by Bipolarity How can I prove this mathematically using the following facts: $P_{1}V_{1} = P_{2}V_{2}$ for a reversible isothermal process $P_{1}V^{γ}_{1} = P_{2}V^{γ}_{2}$ and $γ > 1$ for a reversible adiabatic process. It's really a math problem but I don't know how to begin. BiP
To prove that the adiabatic curve is steeper than the isothermal curve you just need to show that the slope of tangent at a point on the adiabatic curve for a given V is greater than the slope on the other curve with the same value of V.

Or you can do the same by Choosing one value of P and finding the slope of the tangents on the two curves.

I guess you can proceed from here..
Try it.
 P1V1=P2V2 --> P1/P2 = V2/V1 P1V1^c = P2V2^c --> P1/P2 = V2^c/V1^c = (V2/V1)^c Holding P1 and V1 constant and taking c > 1, (V2/V1)^c gets big faster than does (V2/V1), so P2 gets small faster in the second equation than in the first.

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## Isothermal vs Adiabatic expansion

Take the logs of both sides of the equations, and then evaluate what the slopes of log p vs log V would be on a log-log plot.

Chet

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 Quote by Bipolarity Evidently the adiabatic curve is steeper than the isothermal curve. How can I prove this mathematically using the following facts: $P_{1}V_{1} = P_{2}V_{2}$ for a reversible isothermal process $P_{1}V^{γ}_{1} = P_{2}V^{γ}_{2}$ and $γ > 1$ for a reversible adiabatic process. It's really a math problem but I don't know how to begin. BiP
What is dP/dV for each curve?

Write out the equations for each curve with P as a function of V and differentiate with respect to V. Express the slope in terms of T and V (hint: use the ideal gas law).

Is there any point where the slope of the adiabat can have a smaller negative slope than the isotherm for a given V? (hint: what happens if the temperature of the gas in the adiabatic process becomes less than T0/γ? )

AM
 Recognitions: Gold Member Isothermal: (log P2 - log P1)/(log V2 - log V1) = -1 Adiabatic: (log P2 - log P1)/(log V2 - log V1) = - γ γ > 1 Which straight line has a steeper slope on a log-log plot?

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 Quote by Chestermiller Isothermal: (log P2 - log P1)/(log V2 - log V1) = -1 Adiabatic: (log P2 - log P1)/(log V2 - log V1) = - γ γ > 1 Which straight line has a steeper slope on a log-log plot?
Isotherm:

$P = KV^{-1}$

$\frac{dP}{dV} = -KV^{-2} = -P/V$ (substituting K = P/V)

$P = KV^{-\gamma}$

$\frac{dP}{dV} = -\gamma KV^{-(\gamma-1)} = -\gamma P/V$

So which slope is more negative where the two curves have same value of P and V (ie at the beginning)?

You can rewrite these derivatives, substituting P = nRTV^-1:

Isotherm:
$\frac{dP}{dV} =-nRT/V^2$ where T is constant.

$\frac{dP}{dV} = -\gamma nRT/V^2$ where T decreases as V increases.

It seems to me that the slope of the adiabat has to become less negatively sloped than the isotherm when $\gamma T < T_0$, T0 being the temperature of the isotherm.

AM
 Surely you don't need any fancy maths to demonstrate the relationship, a physical explanation of the graph can be made. Both P and V are state functions so ΔP and ΔV are completely determined for any point on the graph. Pick amy two points P1 and P2 with P1 > P2. Then it can be seen from the graph that for any given fall in pressure the concommitent volume increase is greater for an isothermal change than for an adiabtic one. Bipolarity can you offer a physical reason as to why this is so?

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 Quote by Studiot Then it can be seen from the graph that for any given fall in pressure the concommitent volume increase is greater for an isothermal change than for an adiabtic one.
Yes. That is always true. But that doesn't tell you what the relative slopes will be at a given volume.

AM
 If for a given ΔP , ΔViso > ΔVad then ΔViso / ΔP > ΔVad / ΔP But slope = reciprocal of this That is (ΔViso / ΔP)-1 < (ΔVad / ΔP)-1 My question to Bipolarity still stands, as it is designed to help understanding.

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 Quote by Studiot If for a given ΔP , ΔViso > ΔVad then ΔViso / ΔP > ΔVad / ΔP But slope = reciprocal of this That is (ΔViso / ΔP)-1 < (ΔVad / ΔP)-1
Not quite. You run into problems with this analysis in an expansion as P approaches 0. The adiabat may not be able to match the ΔP of the isotherm. So you have to use calculus to find the slope.

The actual slope at any point is dP/dV.

The slope of the isotherm is -nRT/V^2 where T is constant, T0

The slope of the adiabat is -γnRT/V^2 where T is decreasing as V increases.

The OP is asking whether, at any value of V, |dP/dV| is always greater for the adiabat than for the isotherm. It appears to me that at some value of V, γT < T0. Where this occurs the slope of the adiabat will be less negative than the slope of the isotherm.

Looked at another way: if ΔV is the same for both (it is always possible to match ΔV in an expansion) will |ΔP| for the adiabat be greater or less than the |ΔP| for the isotherm? The answer appears to be that if γT > T0 it will be greater and if γT > T0 it will be less.

AM
 If p = 0 you have yourself an honest to goodness vacuum. Nowhere on that PV diagram was P anywhere near zero.
 I find it quite funny how my threads seem to end up. I am glad to be an avid questioner on this site Thank you all for the help. I think I get the point now. BiP
 Andrew wanted a calculus proof. For the isothermal line $$P = \frac{a}{V};\quad {\left( {\frac{{\partial P}}{{\partial V}}} \right)_T} = - \frac{a}{{{V^2}}} = - \frac{a}{V}.\frac{1}{V} = - \frac{P}{V}$$ Similarly for the adiabatic line $$P = \frac{b}{{{V^\lambda }}};\quad {\left( {\frac{{\partial P}}{{\partial V}}} \right)_q} = - \gamma \frac{P}{V}$$ Dividing one by the other $$\frac{{{{\left( {\frac{{\partial P}}{{\partial V}}} \right)}_q}}}{{{{\left( {\frac{{\partial P}}{{\partial V}}} \right)}_T}}} = \gamma$$ For the adiabatic slope to be less than the isothermal would require a gamma of less than 1, but gamma is greater than 1 for ideal gasses. Are there any gases with gamma less than 1? Bipolarity did you answer the question I posed in post #8 - It was designed to help you understand further.

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 Quote by Studiot Andrew wanted a calculus proof. For the isothermal line $$P = \frac{a}{V};\quad {\left( {\frac{{\partial P}}{{\partial V}}} \right)_T} = - \frac{a}{{{V^2}}} = - \frac{a}{V}.\frac{1}{V} = - \frac{P}{V}$$ Similarly for the adiabatic line $$P = \frac{b}{{{V^\lambda }}};\quad {\left( {\frac{{\partial P}}{{\partial V}}} \right)_q} = - \gamma \frac{P}{V}$$ Dividing one by the other $$\frac{{{{\left( {\frac{{\partial P}}{{\partial V}}} \right)}_q}}}{{{{\left( {\frac{{\partial P}}{{\partial V}}} \right)}_T}}} = \gamma$$
That is correct only for one point on the graph - the beginning where the graphs intersect. As V increases, the curves diverge. So for a given V the two Ps are different. So they do not cancel. The ratio of the slopes of these graphs cannot be constant.

This is why mathematics should always be used to check intuition. That is probably why the question in the OP asks for a mathematical proof.

The ratio of the slopes is:

$$\frac{adiabat}{isotherm} = \frac{\gamma\frac{P_a}{V}}{\frac{P_i}{V}} = \gamma\frac{P_a}{P_i} = \gamma\frac{T_a}{T_i}$$

This ratio is not always > 1, which appears to be what you are suggesting.

AM

Well a little play with a spread sheet has convinced me that Andrew is right.

I had always thought that the curves were like in the sketch - they diverge further and further apart, except at infinity.

However in fact the difference between Piso and Pad initially grows but eventually reduces until the curves rejoin at infinity on the Volume axis.

The isothermal curve is always above the adiabatic ie they never cross.

So thank you Bipolarity for posing the question and Andrew for such a good answer. I have learned something here.
Attached Files
 P_V diagram.xls (14.0 KB, 13 views)
 Recognitions: Science Advisor So setting $x=V_2/V_1$ and $y=P_2/P_1$ the first equation reads $$y_\mathrm{it}=x^{-1}$$ and the second one $$y_\mathrm{ad}=x^{-\gamma}$$. The two curves cross at x=y=1, i.e. P_2=P_1 and V_2=V_1. Now $$y'_\mathrm{it}=-x^{-2}$$ and $$y'_\mathrm{ad}=-\gamma x^{-\gamma-1}$$ and $$y'_\mathrm{ad}/ y'_\mathrm{it}=\gamma x^{1-\gamma}.$$ Hence at $x=\gamma^{1/(\gamma-1)}$ the two slopes become equal.