Changes in pressure, temp, & entropy of ideal gas in atmosphere

[baseballfan_ny]

Homework Statement

a) Consider an ideal gas of particles in the earth’s gravitational field,
where each gas molecule has mass m and g is the acceleration due
to gravity. The dependence of the pressure p(z) on the height z is
determined by the condition for mechanical equilibrium: for the
gas contained in a small region, the downward gravitational force
is compensated by the difference between the pressure at the bottom of the region and the pressure at the top. Use this condition
to express dp/dz in terms of m, g, p, and the temperature τ .

b) Suppose that the entropy per particle in the earth’s atmosphere
is independent of altitude, so that $pn^{-\gamma}$
is a constant, where
n = N/V is the concentration, the number of gas molecules per
unit volume. Use the ideal gas law τ = p/n and the result of (a)
to show that dτ /dz is a constant that can be expressed in terms
of γ, m, and g.

Also available as problem #1 on...
http://theory.caltech.edu/~preskill/ph12c/12c-prob4-16.pdf

Relevant Equations

$P = nk_BT$

So for a collection of particles each with mass m, the pressure beneath them, $p(z)$ should be higher than the pressure above them $p(z + \Delta z)$.

This is a change in force per unit area (force per unit volume I suppose) times a volume to equate with the gravitational force
$$ \frac {p(z) - p(z + \Delta z)} {\Delta z} * (\Delta z)^3 = M_{total} * g $$

Introduce the density in the mass
$$ \frac {p(z) - p(z + \Delta z)} {\Delta z} * (\Delta z)^3 = n(z) (\Delta z)^3 g $$

In the limit that $\Delta z$ goes to 0...

$$ -\frac {dp} {dz} = n(z) g $$

Then since it was an ideal gas, $p(z) = n(z)k_B T$

$$ \frac {dp} {dz} = -\frac {p(z)} {k_B T} g $$

Not sure if this is right since my answer doesn't use mass at all (and the question hints at it).

For (b), I'm not sure if I have the background to start yet so I was wondering if I could get some hints on what to look up -- specifically, I don't recognize the $pn^{-\gamma}$ term. I found something online (https://en.wikipedia.org/wiki/Heat_capacity_ratio) about $PV^{-\gamma}$ being constant in adiabatic processes but not sure if this is what the question is looking at.

Also, from class, we derived entropy of particles in an infinite square well (which I believe models an ideal gas) as $$ S = \frac {5} {2} N k_B + Nk_B \ln \left[ \frac {n_Q} {n} \right] $$, $n_Q$ being quantum concentration. So is the problem telling me $\frac {dS} {dz} = 0?$

 

[TSny]

$$ \frac {p(z) - p(z + \Delta z)} {\Delta z} * (\Delta z)^3 = M_{total} * g $$

OK. I guess you are considering a small cube of air of volume $(\Delta z)^3$.

Introduce the density in the mass
$$ \frac {p(z) - p(z + \Delta z)} {\Delta z} * (\Delta z)^3 = n(z) (\Delta z)^3 g $$

The problem statement uses the notation $n$ for the number density, not the mass density. So, $M_{total}$ should involve both $n$ and $m$, where $m$ is the mass of a molecule.

For (b), I'm not sure if I have the background to start yet so I was wondering if I could get some hints on what to look up -- specifically, I don't recognize the $pn^{-\gamma}$ term. I found something online (https://en.wikipedia.org/wiki/Heat_capacity_ratio) about $PV^{-\gamma}$ being constant in adiabatic processes but not sure if this is what the question is looking at.

The correct equation is $PV^\gamma = \rm const$. For a fixed number, $N$, of molecules, you can write $V = N/n$. Thus obtain $Pn^{-\gamma} = \rm const$

Also, from class, we derived entropy of particles in an infinite square well (which I believe models an ideal gas) as $$ S = \frac {5} {2} N k_B + Nk_B \ln \left[ \frac {n_Q} {n} \right] $$, $n_Q$ being quantum concentration. So is the problem telling me $\frac {dS} {dz} = 0?$

For (b), you should only need the two equations $p = n \tau$ and $p/n^\gamma = \rm const$ and your result from (a).

 

[baseballfan_ny]

The problem statement uses the notation n for the number density, not the mass density. So, Mtotal should involve both n and m, where m is the mass of a molecule.

Ah okay... So then my correction is $$\frac {p(z) - p(z + \Delta z)} {\Delta z} * (\Delta z)^3 = n(z) m (\Delta z)^3 g$$

And then in the limit
$$ \frac {dp} {dz} = -\frac {p(z)} {k_B T} mg $$

For (b), you should only need the two equations p=nτ and p/nγ=const and your result from (a).

Alright so I'm just going to switch from $k_B T$ to $\tau$ for no particular reason for this part...

$$ \tau = \frac {p} {n}$$

Now in the next part I kind of assumed $\frac {dn} {dz}$ is 0 and I'm wondering if that's okay to do...
$$ \frac {d \tau} {dz} = \frac {dp} {dz} \frac {1} {n} = \frac {-mg} {k_BT} \frac p n $$

m and g are definitely constant. I don't think I can say T is constant since we're looking to see if its derivative wrt z is constant. I suppose this is where the $p n^{-\gamma}$ relation should come in but I'm not sure how... if I raise n to the $-\gamma$ it would ruin the LHS.

 

[TSny]

Ah okay... So then my correction is $$\frac {p(z) - p(z + \Delta z)} {\Delta z} * (\Delta z)^3 = n(z) m (\Delta z)^3 g$$

And then in the limit
$$ \frac {dp} {dz} = -\frac {p(z)} {k_B T} mg $$

OK.

Alright so I'm just going to switch from $k_B T$ to $\tau$ for no particular reason for this part...

$$ \tau = \frac {p} {n}$$

Now in the next part I kind of assumed $\frac {dn} {dz}$ is 0 and I'm wondering if that's okay to do...

No. Note that the result from (a) tells you that $dp/dz$ is non-zero. So, $p$ varies with $z$. Since $pn^{-\gamma}$ is constant, $n$ must vary with $z$.

The equations that you have to work with are

$p = n \tau$
$p = c n^\gamma , \,\,\,\,\,\,$ $c$ a constant
$dp/dz = -(mgp)/\tau$

I don't know the quickest way to the result. One thing you can try is to eliminate $n$ between the first two equations to get an equation that relates just the two variables $p$ and $\tau$. Take the derivative of both sides of this equation with respect to $z$ to relate $dp/dz$ and $d \tau/dz$. Make use of the third equation in the list above and simplify.

 

[Bystander]

See "Barometric Equation."

 

[baseballfan_ny]

One thing you can try is to eliminate n between the first two equations to get an equation that relates just the two variables p and τ. Take the derivative of both sides of this equation with respect to z to relate dp/dz and dτ/dz. Make use of the third equation in the list above and simplify.

I think I've got something...

I'll call those three eqns you wrote (1), (2), and (3), respectively. Raising (1) to the $\gamma$ and dividing it by (2) gives...
$p^{\gamma - 1} = \frac {\tau^{\gamma}} {c}$

Diffferentiating and applying result from a (Eqn 3)
$$ (\gamma - 1)p^{\gamma - 2} \frac {dp} {dz} = \frac {\gamma \tau^{\gamma - 1} } {c} \frac {d \tau} {dz} $$
$$ (\gamma - 1)p^{\gamma - 2} \frac {-mgp} {\tau} = \frac {\gamma \tau^{\gamma - 1} } {c} \frac {d \tau} {dz} $$
$$ c \frac { (\gamma - 1) } { \gamma } \frac { p^{ \gamma } } { \tau^{\gamma} p^2} \frac {-mgp} {\tau} \tau = \frac {d \tau} {dz} $$
$$ c \frac { (\gamma - 1) } { \gamma } \frac{ n^{\gamma} } {p} \cdot -mg = \frac {d \tau} {dz} $$
$$ c \frac { (\gamma - 1) } { \gamma } \frac{ 1 } {c} \cdot -mg = \frac {d \tau} {dz} $$
$$ \Rightarrow \frac {d \tau} {dz} = \frac { (1 - \gamma) } { \gamma } mg $$

 

[TSny]

Looks correct.

 

[baseballfan_ny]

Thanks all for the help!

See "Barometric Equation."

So physics-wise, all this model is saying is that pressure decreases exponentially with altitude and temperature decreases linearly with altitude (assuming $\gamma$ > 1, which I suppose it would be cause I think it gets colder as you go higher).

I don't really know about adiabatic processes and how the $PV^{\gamma} = c$ came into play here... the question says it's a result of entropy being independent of altitude (I suppose $\frac {dS} {dz} = 0$). Wikipedia says " adiabatic process transfers energy to the surroundings only as work" and none as heat. Is that realistic for the atmosphere and why does it mean $dS = 0$? Is an adiabatic process just a statement that $dS = 0$ in the 1st Law, $dU = TdS - PdV$?

 

[TSny]

Thanks all for the help!So physics-wise, all this model is saying is that pressure decreases exponentially with altitude and temperature decreases linearly with altitude (assuming $\gamma$ > 1, which I suppose it would be cause I think it gets colder as you go higher).

In this "adiabatic model" the pressure does not decrease exponentially according to the "classic" barometric equation $P = P_0 \large e^{-\frac{mgz}{kT}}$. This exponential formula assumes that the temperature is independent of height, which is not the case for the adiabatic model. As an exercise, you can work out $P(z)$ for the adiabatic model using your results so far.

See this link if you want to read more about different thermodynamic models for the atmosphere.

I don't really know about adiabatic processes and how the $PV^{\gamma} = c$ came into play here... the question says it's a result of entropy being independent of altitude (I suppose $\frac {dS} {dz} = 0$). Wikipedia says " adiabatic process transfers energy to the surroundings only as work" and none as heat. Is that realistic for the atmosphere and why does it mean $dS = 0$? Is an adiabatic process just a statement that $dS = 0$ in the 1st Law, $dU = TdS - PdV$?

An adiabatic process is one for which no heat is transferred to or from the system. For a reversible process $dQ = T dS$. So, if $dQ = 0$, then $dS = 0$ for the reversible process. If you google "adiabatic process" you can find lots of reading material.

There really aren't any simple models for the atmosphere that work very well, especially for altitudes higher than about10 km. See the link I gave above. The atmosphere is fairly complicated. I don't know much about it.

 

[TSny]

This recent article shows how Enrico Fermi treated the adiabatic case in his handwritten notes for a course he taught in geophysics.

 

[baseballfan_ny]

See this link if you want to read more about different thermodynamic models for the atmosphere.

Thanks for sharing!