Adiabatic Expansion Pressure Temperature Relation

[SHISHKABOB]

Homework Statement

The problem is in the context of convection in the troposphere

Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation

$$\frac{dT}{dP} = \frac{2}{f+2} \frac{T}{P}$$

Homework Equations

Ideal gas law PV = nRT
adiabatic relations $VT^{f/2}= constant$, $V^{\gamma}P = constant$
where $\gamma = \frac{f+2}{f}$

The Attempt at a Solution

The first thing I tried was taking the derivative of each of the adiabatic relations w.r.t. the T or P in each of them. Since they are equal to constants, their derivatives are equal to zero and thus each other. I did not take a partial derivative because V is not constant. Temperature, pressure and volume of the air mass are changing as it moves through the atmosphere.

I am rusty with my derivatives right now, and was plagued with simple errors as I went through taking them and doing the arithmetic. Like right now I see a mistake that I made that I didn't notice I had and yeah, I gave up on this route because it was too messy.

Then I tried the idea of dividing the one adiabatic relation by the other, and since the ratio of two constants is a constant, I could then take the derivative of what I got and work from there.

$$PV^\gamma = A$$
$$VT^{f/2} = B$$
$$\frac{PV^\gamma}{VT^{f/2}} = A/B$$
$$PV^{\gamma - 1}T^{-f/2} = constant$$

From there I uh wrote the "implicit derivative" of each. I'm not entirely sure if this is legal, or if I did it right but

$$V^{\gamma - 1}T^{-f/2}dP + PT^{-f/2}V^{\gamma - 2}(\gamma - 2)dV + PV^{\gamma - 1}T^{-f/2 - 1}(-f/2 - 1)dT = 0$$

which is admittedly still messy but whatever. I moved all the Vs to the one side of the equation to leave the Ps and Ts on the other

$$\frac{dP}{P} = (f/2 + 1) \frac{dT}{T} - (\gamma - 2)\frac{dV}{V}$$

From an example in the book I was shown that $\frac{f}{2}\frac{dT}{T} = -\frac{dV}{V}$. I'm assuming this is true for my problem because the example was about adiabatic compression and so P, V and T were dependent on each other like in the problem I'm doing.

And so $$\frac{dP}{P} = \frac{dT}{T}[(\gamma -2)(f/2) + (f/2 + 1)]$$

so now I'm going to simplify and turn the $\gamma$ into $(f+2)/f$and holy crap if I didn't redo this a million times but I keep ending up with

$$\frac{dP}{P} = \frac{f+4}{2}\frac{dT}{T}$$

Which is exactly the answer (with some maneuvering) except there's a FOUR where a TWO should be!

Ok I found that I was writing a 1 where the 2 should be in $\gamma - 2$... which makes the above turn out to be $$\frac{dP}{P} = 2\frac{dT}{T}$$

I can only pore over my handiwork so many times looking for tiny errors until my mind starts to turn off, but I can't help but think I did something stupid. Either that or perhaps that substitution is fallacious? I tried changing the sign of that part: $\frac{f}{2}\frac{dT}{T} = \frac{dV}{V}$ (maybe because the work is in the opposite direction for expansion) but it did not help either.

And ok at this point I'm just confusing myself with all my tiny changes and am going to go take a break. I am very appreciative of any help, thank you.

 

[Chestermiller]

Start with PV^γ=constant, and substitute V from the ideal gas law into the equation. What do you get?

 

[SHISHKABOB]

I get $$\frac{dT}{dP} = \frac{\gamma}{\gamma - 1} \frac{T}{P^{\gamma + 1}}$$

Which evaluates to $$\frac{dT}{dP} = \frac{f+2}{2} \frac{T}{P^{\frac{2f+2}{f}}}$$ and I really don't know how to proceed from there.

 

[SHISHKABOB]

Oh wow wait I think I got it, I took the natural log of the initial substitution and I think that's exactly what I needed to do. I can nicely separate P and T from each other using log properties, and no weird exponents.

I could take the log, of course, because a ln(constant) is still some constant.

 

[Chestermiller]

I get $$\frac{dT}{dP} = \frac{\gamma}{\gamma - 1} \frac{T}{P^{\gamma + 1}}$$

That's not what I get. I get:
$$T=CP^{\frac{(\gamma-1)}{\gamma}}$$
where C is a constant.

 

[SHISHKABOB]

Thanks, that did it!

 

[Lowell]

For future people.

Use
\begin{equation}
VT^{ \frac {f} {2}} = C
\end{equation}
and
\begin{equation}
V= \frac {NKT} {P}
\end{equation}
Enter Equation (2) into Equation (1)
\begin{equation}
( \frac {NKT} {P})T^{\frac {f} {2}}=C
\end{equation}
Combine T's and rearrange to be a function of T
\begin{equation}
T^ {\frac {f+2} {2}}= \frac {CP} {NK}
\end{equation}
Recognize that
\begin{equation}
\frac {C} {NK} = constant
\end{equation}

Enter that information
\begin{equation}
T^{ \frac {f+2}{2} }= CP
\end{equation}
Multiply by exponential \begin{equation} ( \frac {2} {f+2} ) \end{equation} to isolate a single T. (C raised to any power is still a constant).
\begin{equation}
T=CP^{ \frac {2} {f+2}}
\end{equation}
Now take the derivative of T with respect to P.
\begin{equation}
\frac {dT} {dP} = \frac {2C} {f+2} P^{\frac {2} {f+2} -1}
\end{equation}
separate P^-1 from P^2/f+2
\begin{equation}
\frac {dT} {dP} = \frac {2C} {f+2} P^{\frac {2} {f+2}}P^{-1}
\end{equation}
Rearrange (
\begin{equation}
\frac {T} {C} = P^{\frac {2} {f+2}}
\end{equation}
Input above equation into the differential for matching P.
\begin{equation}
\frac {dT} {dP} = \frac {2C} {f+2} (\frac {T} {C})P^{-1}
\end{equation}
Constants are equal so they can cancel out. Leaving you with
\begin{equation}
\frac {dT} {dP} = \frac {2} {f+2} \frac {T} {P}
\end{equation}

 

[Chestermiller]

For future people.

Use
\begin{equation}
VT^{ \frac {f} {2}} = C
\end{equation}
and
\begin{equation}
V= \frac {NKT} {P}
\end{equation}
Enter Equation (2) into Equation (1)
\begin{equation}
( \frac {NKT} {P})T^{\frac {f} {2}}=C
\end{equation}
Combine T's and rearrange to be a function of T
\begin{equation}
T^ {\frac {f+2} {2}}= \frac {CP} {NK}
\end{equation}
Recognize that
\begin{equation}
\frac {C} {NK} = constant
\end{equation}

Enter that information
\begin{equation}
T^{ \frac {f+2}{2} }= CP
\end{equation}
Multiply by exponential \begin{equation} ( \frac {2} {f+2} ) \end{equation} to isolate a single T. (C raised to any power is still a constant).
\begin{equation}
T=CP^{ \frac {2} {f+2}}
\end{equation}
Now take the derivative of T with respect to P.
\begin{equation}
\frac {dT} {dP} = \frac {2C} {f+2} P^{\frac {2} {f+2} -1}
\end{equation}
separate P^-1 from P^2/f+2
\begin{equation}
\frac {dT} {dP} = \frac {2C} {f+2} P^{\frac {2} {f+2}}P^{-1}
\end{equation}
Rearrange (
\begin{equation}
\frac {T} {C} = P^{\frac {2} {f+2}}
\end{equation}
Input above equation into the differential for matching P.
\begin{equation}
\frac {dT} {dP} = \frac {2C} {f+2} (\frac {T} {C})P^{-1}
\end{equation}
Constants are equal so they can cancel out. Leaving you with
\begin{equation}
\frac {dT} {dP} = \frac {2} {f+2} \frac {T} {P}
\end{equation}

You realize that this thread is over a year old, right? Do you feel that the answer we previously obtained was incorrect?

 

[Lowell]

You realize that this thread is over a year old, right? Do you feel that the answer we previously obtained was incorrect?

In this problem you aren't searching for an answer. You are trying to figure out the necessary steps for derivation. So I was just putting this here for future people searching for it.

 

[Chestermiller]

In this problem you aren't searching for an answer. You are trying to figure out the necessary steps for derivation. So I was just putting this here for future people searching for it.

Well, OK. Since the OP has not been seen in over a year, and there have been no other responses until yours, I think this thread has pretty much run its course. I am hereby closing it.