Correct Derivation of the Adiabatic Condition? (PV Diagram)

[PhDeezNutz]

Homework Statement

Derive the condition the Adiabatic Condition $PV^{\gamma} = \text{constant}$ for a monatomic ideal gas. I believe $\gamma = \frac{5}{3}$.

Relevant Equations

Equipartition Theorem:
$E = \frac{3}{2} NkT \Rightarrow \Delta E = \frac{3}{2} Nk \Delta T$

Ideal Gas Law:

$PV = NkT \Rightarrow P \Delta V + V \Delta P = Nk \Delta T$

First Law:

$\Delta E = Q - W$

Adiabatic:

$Q = 0$

By the First Law, Definition of an Adiabatic Process, and Definition of Work:

$\Delta E = Q - W = - W = - P \Delta V$ (because $Q = 0$) (Equation 1)

By the Equipartition Theorem:

$\Delta E = \frac{3}{2} Nk \Delta T$ (Equation 2)

By Setting Equation 1 equal to Equation 2

$\Delta T = - \frac{2P}{3Nk} \Delta V$ (Equation 3)

Differential Form of Ideal Gas Law:

$P \Delta V + V \Delta P = Nk \Delta T$ (Equation 4)

Plug Equation 3 into Equation 4


$P \Delta V + V \Delta P = Nk \Delta T = - \frac{2P}{3} dV$ (Equation 5)

So getting rid of the middle man

$P \Delta V + V \Delta P = - \frac{2P}{3} dV$ (Equation 6)

Moving the LHS over to the RHS we have


$\frac{5}{3} P \Delta V + V \Delta P = 0$ (Equation 7)

This can be stated another way


$\frac{\Delta P}{\Delta V} + \frac{5}{3} \left( \frac{1}{V} \right) P = 0$ (Equation 8)


Equation 8 can be re-written in a calculus friendly way and we can use some basic methods of Differential Equations to solve for $P$ as a function of $V$ or just stop a step before doing that and establish $PV^{\gamma} = \text{constant}$

$\frac{ dP}{dV} + \frac{5}{3} \left( \frac{1}{V} \right) P = 0$ (Equation 8)

multiply through by integrating factor $\mu = e^{\int \frac{5}{3V} \,dV} = e^{ln v^{\frac{5}{3}}} = v^{\frac{5}{3}}$

$v^{\frac{5}{3}} \frac{ dP}{dV} + \frac{5}{3} \left(v^{\frac{2}{3}} \right) P = 0$ (Equation 8)

We recognize the LHS as the derivative of $PV^{\frac{5}{3}}$ So

$\frac{d}{dV} \left( PV^{\frac{5}{3}}\right) = 0$ (Equation 9)

So

$PV^{\frac{5}{3}} = \text{constant}$ (Equation 10)

of course if we really wanted to we could say

$P = \frac{\text{constant}}{V^{\frac{5}{3}}}$

Hopefully I didn't play too fast and loose with derivatives $d$ and full changes $\Delta$, thanks in advance for any help/guidance.

If the above work is correct, how can we establish that "Adiabatic process on the PV diagram are steeper than those of isotherms".............after all we don't necessarily know what each "constant" is in

$PV = \text{constant}_1$ (Isotherm)

$PV^{\frac{5}{3}} = \text{constant}_2$ (Adiabat)

Are $\text{constant}_1$ and $\text{constant}_2$ the same?

 

[PhDeezNutz]

Maybe this is too simplistic but maybe it will get some ideas rolling

if we have two parallel isotherms (they have to be parallel at least locally for a given $V$) to get from one to the other $T$ has to change. The only way to get from one $T$ to another $T$ (i.e. from one isotherm to another) is to follow a path that connects the two isotherms which is necessarily steeper than either isotherm.

 

[haruspex]

Are $\text{constant}_1$ and $\text{constant}_2$ the same?

It is meaningless to ask if they have the same value since they are dimensionally different.
The claim that one is steeper then the other would make sense on a log-log chart.

 

[PhDeezNutz]

It is meaningless to ask if they have the same value since they are dimensionally different.
The claim that one is steeper then the other would make sense on a log-log chart.

Fair but what about my reasoning in post 2?:

If isotherms are parallel and in adiabatic processes $T$ changes then adiabats must "travel between" isotherms making them necessarily steeper?

I will ponder your log-log argument and respond.

 

[haruspex]

Fair but what about my reasoning in post 2?:

If isotherms are parallel and in adiabatic processes $T$ changes then adiabats must "travel between" isotherms making them necessarily steeper?

That's probably valid, but you can also get there by writing down the equations for how P depends on V in the two cases, taking the derivatives, evaluating at $(P_0, V_0)$and taking the ratio.