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

In summary: BT} \frac p n $$which is equal to$$ \frac {dS} {dz} = \frac {p(z) - p(z + \Delta z)} {\Delta z} * (\Delta z)^3 = \frac {-mg} {k_BT} \frac p n = 0$$
  • #1
baseballfan_ny
92
23
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? ##
 
Physics news on Phys.org
  • #2
baseballfan_ny said:
$$ \frac {p(z) - p(z + \Delta z)} {\Delta z} * (\Delta z)^3 = M_{total} * g $$
OK. I guess you are considering a small cube of air of volume ##(\Delta z)^3##.

baseballfan_ny said:
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.

baseballfan_ny said:
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##

baseballfan_ny said:
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).
 
  • Like
Likes baseballfan_ny
  • #3
TSny said:
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
$$ \frac {dp} {dz} = -\frac {p(z)} {k_B T} mg $$

TSny said:
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.
 
  • #4
baseballfan_ny said:
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.

baseballfan_ny said:
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.
 
  • Like
Likes baseballfan_ny
  • #5
See "Barometric Equation."
 
  • Like
Likes baseballfan_ny
  • #6
TSny said:
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)
$$ (\gamma - 1)p^{\gamma - 2} \frac {dp} {dz} = \frac {\gamma \tau^{\gamma - 1} } {c} \frac {d \tau} {dz} $$
$$ (\gamma - 1)p^{\gamma - 2} \frac {-mgp} {\tau} = \frac {\gamma \tau^{\gamma - 1} } {c} \frac {d \tau} {dz} $$
$$ c \frac { (\gamma - 1) } { \gamma } \frac { p^{ \gamma } } { \tau^{\gamma} p^2} \frac {-mgp} {\tau} \tau = \frac {d \tau} {dz} $$
$$ c \frac { (\gamma - 1) } { \gamma } \frac{ n^{\gamma} } {p} \cdot -mg = \frac {d \tau} {dz} $$
$$ c \frac { (\gamma - 1) } { \gamma } \frac{ 1 } {c} \cdot -mg = \frac {d \tau} {dz} $$
$$ \Rightarrow \frac {d \tau} {dz} = \frac { (1 - \gamma) } { \gamma } mg $$
 
  • Like
Likes TSny
  • #7
Looks correct.
 
  • Like
Likes Delta2
  • #8
Thanks all for the help!

Bystander said:
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##?
 
  • #9
baseballfan_ny said:
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.

baseballfan_ny said:
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.
 
  • Like
Likes baseballfan_ny
  • #10
This recent article shows how Enrico Fermi treated the adiabatic case in his handwritten notes for a course he taught in geophysics.
 
  • Like
Likes baseballfan_ny
  • #11
TSny said:
See this link if you want to read more about different thermodynamic models for the atmosphere.
Thanks for sharing!
 

1. What is the ideal gas law and how does it relate to changes in pressure, temperature, and entropy of gases in the atmosphere?

The ideal gas law is a mathematical equation that describes the relationship between pressure, volume, temperature, and the number of moles of a gas. It can be written as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. This law helps us understand how these variables change when a gas is heated or cooled, and how changes in pressure can affect the volume and temperature of a gas.

2. How do changes in pressure affect the behavior of an ideal gas in the atmosphere?

Changes in pressure can affect the volume and temperature of an ideal gas in the atmosphere. According to the ideal gas law, if the pressure of a gas increases, its volume decreases, and vice versa. This is because the molecules of a gas are constantly moving and colliding with each other and the walls of their container. An increase in pressure means more collisions, which leads to a decrease in volume. Similarly, a decrease in pressure means fewer collisions and an increase in volume.

3. How does temperature impact the behavior of an ideal gas in the atmosphere?

Temperature plays a crucial role in the behavior of an ideal gas in the atmosphere. As temperature increases, the average kinetic energy of the gas molecules also increases, causing them to move faster and collide more frequently with each other and the walls of their container. This results in an increase in pressure and volume. Conversely, a decrease in temperature leads to a decrease in kinetic energy and a decrease in pressure and volume.

4. What is entropy and how does it relate to changes in pressure and temperature of an ideal gas in the atmosphere?

Entropy is a measure of the disorder or randomness of a system. In the case of an ideal gas in the atmosphere, an increase in pressure or temperature leads to an increase in entropy. This is because an increase in pressure or temperature causes the gas molecules to move more randomly and occupy a larger space, resulting in a higher degree of disorder.

5. How do changes in pressure, temperature, and entropy of an ideal gas in the atmosphere affect weather patterns and climate?

The changes in pressure, temperature, and entropy of gases in the atmosphere play a significant role in shaping our weather patterns and overall climate. For example, changes in pressure can lead to the formation of high and low-pressure systems, which can influence wind patterns and the movement of weather systems. Changes in temperature can also affect the amount of moisture in the air, leading to changes in precipitation patterns. Additionally, changes in entropy can impact the overall stability of the atmosphere and contribute to the formation of severe weather events, such as hurricanes and tornadoes.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
739
  • Introductory Physics Homework Help
Replies
2
Views
648
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
998
  • Introductory Physics Homework Help
Replies
1
Views
631
  • Introductory Physics Homework Help
Replies
32
Views
962
  • Introductory Physics Homework Help
Replies
10
Views
1K
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
1K
Back
Top