Physics Quiz Question

[Dallas]

Brought over from RAS:


Assuming that all other variables remain constant:

An increase in temperature will result in a higher atmospheric pressure - a
higher temperature speeds up the movement of the air molecules, thereby
raising the pressure they exert on the surrounding atmosphere.

A) True
B) False

--
Dallas

[Paul Tomblin]

In a previous article, said:
>Brought over from RAS:
>
>
>Assuming that all other variables remain constant:
>
>An increase in temperature will result in a higher atmospheric pressure - a
>higher temperature speeds up the movement of the air molecules, thereby
>raising the pressure they exert on the surrounding atmosphere.
>
>A) True
>B) False

That would be true if the atmosphere was a closed container. It's not,
it's free to expand, so the pressure does not rise.


--
Paul Tomblin > http://blog.xcski.com/
Never underestimate the bandwidth of a station wagon full of
tapes hurtling down the highway.
-- Andrew Tannenbaum possibly quoting Warren Jackson

[alexy]

(Paul Tomblin) wrote:

>In a previous article, said:
>>Brought over from RAS:
>>
>>
>>Assuming that all other variables remain constant:
>>
>>An increase in temperature will result in a higher atmospheric pressure - a
>>higher temperature speeds up the movement of the air molecules, thereby
>>raising the pressure they exert on the surrounding atmosphere.
>>
>>A) True
>>B) False
>
>That would be true if the atmosphere was a closed container. It's not,
>it's free to expand, so the pressure does not rise.

I think you are right, but is it a trick question? He said "all other
variables remain constant" Does that include the volume of the
atmosphere? Never mind that such an assumption makes no sense.

--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

[Dallas]

On Mon, 06 Aug 2007 14:34:07 -0400, alexy wrote:

> I think you are right, but is it a trick question?

It's not supposed to be a tricky question. I posted it because everyone
seems to have a different answer for it. It's a tiny bit like the old
"airplane on a treadmill" debate.

--
Dallas

[Doug Semler]

On Aug 6, 1:05 pm, Dallas > wrote:
> Brought over from RAS:
>
> Assuming that all other variables remain constant:
>
> An increase in temperature will result in a higher atmospheric pressure - a
> higher temperature speeds up the movement of the air molecules, thereby
> raising the pressure they exert on the surrounding atmosphere.
>
> A) True
> B) False
>

Well, the answer is false but it is a rather bogus problem, mainly
because the atmosphere is not fully contained; in other words, the
volume CAN change. The way this is worded, the fluid (atmospheric gas)
is also being used as the container. Pressure is a measurement of
normal (perpendicular) force against a surface, you can't have a
pressure without a surface against which an equal and opposite
pressure can be exerted. Since the atmosphere can expand (because it
is not closed), it doesn't have anything to exert pressure against
EXCEPT for the earth's surface.

But I invite you to think about: Of what is a temperature a
measurement, how does force relate to potential energy and mass, how
pressure relates to force and mass, how potential and kinetic energy
is related, how kinetic energy is related to velocity, what the laws
of conservation of energy entail, and what "atmospheric pressure" is).

Hint: "atmospheric pressure" is a measurement of the pressure of the
air due to the acceleration of gravity on the mass of column of air
directly above it, not as a measure of pressure against the
"surrounding" atmosphere.

[Capt. Geoffrey Thorpe]

"Dallas" > wrote in message
...
> Brought over from RAS:
>
>
> Assuming that all other variables remain constant:
>
> An increase in temperature will result in a higher atmospheric pressure -
> a
> higher temperature speeds up the movement of the air molecules, thereby
> raising the pressure they exert on the surrounding atmosphere.
>
> A) True
> B) False

False and/or True.

The air can expand. If it expands vertically, the pressure is unchanged. If
it expands horizantally, the pressure goes down. If the air is constrained,
the pressure goes up.

Depends on what you mean by "all other variables remain constant"

PV = nRT

--
Geoff
The Sea Hawk at Wow Way d0t Com
remove spaces and make the obvious substitutions to reply by mail
When immigration is outlawed, only outlaws will immigrate.

[Tina]

the variable that matters is that there's 14.7 pounds of air over
every square inch at sea level. If that's the one that stays constant,
the answer is clear.

On Aug 6, 3:00 pm, "Capt. Geoffrey Thorpe" <The Sea Hawk at wow way
d0t com> wrote:
> "Dallas" > wrote in message
>
> ...
>
> > Brought over from RAS:
>
> > Assuming that all other variables remain constant:
>
> > An increase in temperature will result in a higher atmospheric pressure -
> > a
> > higher temperature speeds up the movement of the air molecules, thereby
> > raising the pressure they exert on the surrounding atmosphere.
>
> > A) True
> > B) False
>
> False and/or True.
>
> The air can expand. If it expands vertically, the pressure is unchanged. If
> it expands horizantally, the pressure goes down. If the air is constrained,
> the pressure goes up.
>
> Depends on what you mean by "all other variables remain constant"
>
> PV = nRT
>
> --
> Geoff
> The Sea Hawk at Wow Way d0t Com
> remove spaces and make the obvious substitutions to reply by mail
> When immigration is outlawed, only outlaws will immigrate.

[terry]

On Aug 7, 3:05 am, Dallas > wrote:
> Brought over from RAS:
>
> Assuming that all other variables remain constant:
>
> An increase in temperature will result in a higher atmospheric pressure - a
> higher temperature speeds up the movement of the air molecules, thereby
> raising the pressure they exert on the surrounding atmosphere.
>
> A) True
> B) False
>
> --
The answer is true, but the usefulness of the question is doubtfall
because all things dont remain constant. In practice , at least where
I live, the temperature can change quite radily during the day with
virtually no change in pressure. But then when you talk about
synoptic charts, high pressure is associated with higher temperatures,
and vice versa for low pressure systems.
terry

[Doug Semler]

On Aug 6, 9:37 pm, Tina > wrote:
> the variable that matters is that there's 14.7 pounds of air over
> every square inch at sea level. If that's the one that stays constant,
> the answer is clear.
>

Well...not to quibble, but 14.7 (ok, 14.694) psi is the MEAN sea level
pressure. There are localized variations in pressure caused by
density differences over those areas. (which is why GENERALLY
speaking, areas of lower pressure indicate "rain" systems; the lighter
air indicates a higher moisture content).

Although, in large enough volumes the differences are close enough to
zero that they do not significantly affect calculations since

lim (m/V) = 0
V->inf

[Dallas]

On Mon, 6 Aug 2007 12:05:17 -0500, Dallas wrote:

> An increase in temperature will result in a higher atmospheric pressure - a
> higher temperature speeds up the movement of the air molecules, thereby
> raising the pressure they exert on the surrounding atmosphere.
> A) True > B) False

The origin of the question comes from the current edition of the Jeppesen
Private Pilot manual.

Here's the exact text:
http://img.photobucket.com/albums/v101/Dallas52/Dallas/Jep1.jpg

Which would seem to imply that the answer is true.

If you're wondering if it's out of context, here's all the text on the
page:

http://img.photobucket.com/albums/v101/Dallas52/Dallas/Jep2.jpg


--
Dallas

[Jim Logajan]

Dallas > wrote:
> The origin of the question comes from the current edition of the
> Jeppesen Private Pilot manual.
[...]
>
> http://img.photobucket.com/albums/v101/Dallas52/Dallas/Jep2.jpg

Context helps. The kicker, and misleading, aspect is this:

"...assuming that all other variables remain constant...."

First, what the Jeppesen manual states is self-evident from the ideal gas
law:

P*V = n*R*T

So their text is technically correct, but really incorrect as far as
describing what *really* happens when some part of the atmosphere heats
up more than the surrounding atmosphere.

In the real atmosphere the volume V0 that is occupied by some n moles of
gas does not "remain constant". But neither does a volume V0 of gas
undergo "free expansion" as the temperature rises from T0 to T1, as I've
seen some people suggest. To be clear: "free expansion" means the total
energy of the n molecules in V0 remains the same as the volume goes to V1
- that is, as the gas expands it does no work, and no work is done on it.
(Or put mathematically, "free expansion" says that P0*V0 = P1*V1; that is
because the product of pressure and volume yields units of energy!) Of
course it is not the case that the energy remains constant because as the
gas expands it has to do work against gravity (i.e. it has to push
against the surrounding atmosphere). So in general what actually happens
is something _in between_ what happens when V0 = V1 (volume held
constant) and P0*V0 = P1*V1 (energy held constant). (And to determine P1
and V1 is why thermodynamics texts are filled with imposing looking
differential equations! :-))

BUT THE BEST FORMULA (and most relevant to aviators) DOESN'T DEAL WITH
VOLUME CHANGES! IT DEALS WITH DENSITY CHANGES! Here is how it is derived:

First divide both sides of the ideal gas law by V and rearrange
variables:

P = R*T*(n/V)

But n/V is just a density! So density = n/V and we get:

P = R*T*density

But R is "just" a conversion constant to make sure all the units work
out. If we only want to understand proportionalities we can discard the
R. Then dividing both sides by T yields:

density = P/T

*** SO IMHO THE BEST FORMULATION OF THE IDEAL GAS LAW FOR PILOTS IS: ***
**** ****
****** DENSITY = PRESSURE/TEMPERATURE ******

So when Jeppesen said "assuming all other variables remain constant" it
was basically saying "assuming the density remains constant." But air
density is the ultimate variable of interest to pilots! So Jeppesen was
effectively posing an example that said "assuming pressure and
temperature vary such that it doesn't affect the lift produced by your
wings!" Now *that* is a misleading (and useless) example IMHO!

[Dallas]

On Tue, 07 Aug 2007 18:31:41 -0000, Jim Logajan wrote:

> So when Jeppesen said "assuming all other variables remain constant" it
> was basically saying "assuming the density remains constant."

Are you pretty comfortable with that statement? I can't imagine why
Jeppesen would bother to publish the paragraph if the assumption was that
the density would remain constant, which is basically impossible outside
the laboratory, (at least, as far as I know) and makes whole statement of
no value to a pilot in the real world.

"Assuming all other variables remain constant". - I picture two barometers
a few miles apart on a consistent, flat surface. There is no wind and the
sky is overcast. A hole in the overcast opens up and heats the area around
the first barometer. If I correctly interpret what Jeppesen appears to be
saying, the pressure in the area of the heated barometer will rise above
the barometer in the shade.


--
Dallas

[Doug Semler]

On Aug 7, 3:32 pm, Dallas > wrote:
> On Tue, 07 Aug 2007 18:31:41 -0000, Jim Logajan wrote:
> > So when Jeppesen said "assuming all other variables remain constant" it
> > was basically saying "assuming the density remains constant."
>
> Are you pretty comfortable with that statement? I can't imagine why
> Jeppesen would bother to publish the paragraph if the assumption was that
> the density would remain constant, which is basically impossible outside
> the laboratory, (at least, as far as I know) and makes whole statement of
> no value to a pilot in the real world.
>
> "Assuming all other variables remain constant". - I picture two barometers
> a few miles apart on a consistent, flat surface. There is no wind and the
> sky is overcast. A hole in the overcast opens up and heats the area around
> the first barometer. If I correctly interpret what Jeppesen appears to be
> saying, the pressure in the area of the heated barometer will rise above
> the barometer in the shade.

OK, I think I understand what's going on. You are interpreting
something that, while technically correct, is aimed at non-physists.
All it is trying to say is that if you have the same mass of air
contained in the same volume, but with the air at different
temperatures, the pressures will be different. One of the main issues
of contention that I have is that we are talking about energy
transfers that affect multiple different component variables (e.g.
mass, volume, and temperature) in an open system, and trying to close
the system AND hold all but one of them constant (the old "ignoring
friction" routine).

I still don't like their use of "exerting pressure" on the surrounding
atmosphere, because I really don't think it is. However, I guess you
could demonstrate their statement by using an infinitely thin, capped,
and flexible column surrounding a mass of air (think condom shape).
if you heated the air inside of it, you would see the column walls bow
to the pressure changes. But the walls bowing is a demonstration of
the pressure differential and attempt to establish equalibrium more
than anything else, and handwaves the fact that pressure isn't defined
this way.

I think Jim is right; a more important relation is how a particular
temperature reading at a particular pressure reading relates to the
density of the air at altitude. This is probably one of those cases
where <cursory hand wave...it's close enough> about how it works is
less important than the effects that it has on an aircraft's
performance at a given altitude for different conditions.

(see also further example in r.a.s.)

[Jim Logajan]

Dallas > wrote:
> On Tue, 07 Aug 2007 18:31:41 -0000, Jim Logajan wrote:
>
>> So when Jeppesen said "assuming all other variables remain constant"
>> it was basically saying "assuming the density remains constant."
>
> Are you pretty comfortable with that statement?

Pretty comfortable. Although I admit I did better in quantum mechanics in
college than thermo and statistical phsyics. :-)

> I can't imagine why
> Jeppesen would bother to publish the paragraph if the assumption was
> that the density would remain constant, which is basically impossible
> outside the laboratory, (at least, as far as I know) and makes whole
> statement of no value to a pilot in the real world.

Well, it looks like they were trying to make a point and be technically
accurate. I believe that typically requires making text-book simplifying
assumptions.

> "Assuming all other variables remain constant". - I picture two
> barometers a few miles apart on a consistent, flat surface. There is
> no wind and the sky is overcast. A hole in the overcast opens up and
> heats the area around the first barometer. If I correctly interpret
> what Jeppesen appears to be saying, the pressure in the area of the
> heated barometer will rise above the barometer in the shade.

Your interpretation matches exactly what they say: "On the other hand, a
warmer temperature increases atmospheric pressure, all else being equal."

But the problem is that "all else" _doesn't_ remain equal in your
scenario. That is why I think the Jeppesen paragraph is misleading, since
it will lead students to believe certain values remain constant when they
really don't.

In your scenario the air over the sunny area may try to increase in
pressure to, for example, 14.8 lb/in^2 with surrounding air at, say, 14.7
lb/in^2. The pressure difference will cause the heated air to balloon
outward till the pressures at the imaginary boundary equalize. What you
get is an outflowing "wind" as the hotter air balloons out. The hot area
should also cool a little. The details get ugly, but suffice to say that
almost immediately after the heating begins, the volume starts expanding
so the density starts dropping and the barametric pressure will appear,
in this example, to be between 14.7 and 14.8.

So in general if it is going to be a hot day, the air density will _tend_
to be either the same or less than on a cooler day. So wing lift and
oxygen content/intake is slightly reduced on the hotter day.

But dang it, nothing in physics seems to rule out weather events
conspiring so that one gets a hot day and a high pressure system such
that the air density is much higher than average. Such is the nature of a
dynamic atmosphere that experiences unequal heating.

[Jim Logajan]

Clark > wrote:
> A minor nitpick on a previous post from Jim:
>
> It was stated that n/V was equal to density.

A reasonable nit, but I was careful enough to say it was "a" density, not
"the" density. My words were: "But n/V is just a density!"

> While the stated equality
> isn't, the jist of what was said was ok. The equality would be
>
> (Mass/Molecular Weight)/V = density
>
> Since the molecular weight of air is a constant it can be combined
> with the other constant in the ideal gas law when used for atmospheric
> calcs.

Thanks for the nit pick and clarification.

Nowadays I sometimes qualify things that don't need qualification, and fail
to qualify things I should. For example, you might find me stating "It is
my understanding that 1 + 1 = 2 for most values of 1 and 2," rather than
the unequivical "1 + 1 = 2". Otherwise somebody will go:

"Bzzzzt! Wrong! Try again. 1 + 1 = 3. And here's the proof...."

And by gosh the proof would look valid. Or they'd direct me to a .gov web
page where it states that 1 + 1 = 3. And since I like to use as
"authoritative" web sites as I can find to support my own assertions, and I
consider .gov sites reasonably authoritative, I'd be forced to admit my
ignorance and stupidity for making such an unqualified assertion. It
becomes tiring admitting to ignorance and stupidity too often (at least it
is for me!), and I suspect that after a while the only people left reading
my posts are those who are equally ignorant and stupid. At that point I
think my posting becomes futile/redundant since my ignorance and stupidity
has reached equilibrium with the equally ignorant and stupid readers of my
posts.

What, me ramble?

[Tony]

Think of it this way. If the entire atmosphere's temperature was
increased by say 10 degrees, the average pressure at the surface would
be as it had been, each square inch supporting about 15 pounds of air.
The 15 pounds doesn't change bcause it's hotter.



On Aug 6, 10:05 am, Dallas > wrote:
> Brought over from RAS:
>
> Assuming that all other variables remain constant:
>
> An increase in temperature will result in a higher atmospheric pressure - a
> higher temperature speeds up the movement of the air molecules, thereby
> raising the pressure they exert on the surrounding atmosphere.
>
> A) True
> B) False
>
> --
> Dallas

[Doug Semler]

On Aug 8, 12:25 am, Clark > wrote:
> Jim Logajan > wrote :
>
> > Clark > wrote:
> >> A minor nitpick on a previous post from Jim:
>
> >> It was stated that n/V was equal to density.
>
> > A reasonable nit, but I was careful enough to say it was "a" density,
> > not "the" density. My words were: "But n/V is just a density!"
>
> Nit picking my nitpic? Hmmm, is anyone keeping score? :-) You are correct of
> course but maybe we can take a closer look. Lets-see-here-now, we have
> density in moles/L^3. Is that a nuclear density? Whatever it is it must be a
> Chem E thing... I've seen flow rates in moles/hour, but this is new. Ya never
> can trust those Chem E types anyways.
>

Nope, ya just can't. What the hell is a cubic liter anyway? I've
heard of cubic meters, but never cubic liters <g>. And just when you
think your all smart, they start throwing moles at you. Sheesh.

Seriously, though, I too have a tendency to skip steps/simplify things
when doing this sort of thing. The target audience here is not a
bunch of physicists, it's a bunch of pilots. While some of them might
care/want to know how it works in a general sense, it seems silly to
start bringing out all the math involved, and easier to make some wild
ass assumptions that are only valid in a theoretical model. The silly
theoretical model can help explain what's going on, and may help
someone understand that temperature, pressure, and density are all
related to each other. I try to indicate when I am <hand waving> a
"that difference is close enough to zero to not matter", but I
sometime forget. However, problems like this is like the high school
physics problems that "ignore friction". Understanding that friction
DOES play a part of an overall system is important; the principles
being taught are MORE important. (little delta/epsilon are implied).

I admit I am not a physicist, I *am* a mathemetician, and some upper
division calculus type classes are very closly linked with physics;
hell a whole class was devoted to the calculus of thermodynamics.
Most of the stuff I am writing is coming from memory 10 years old or
more, so it probably does have some holes in it, if I am wrong i am
sure to be corrected (as I was in r.a.s; i fell into the trap of not
fully explaining my meaning <sigh>). But that doesn't change the fact
that I don't think anyone in *this* group wants me to dust off the 'ol
applied calculus book and start quoting dry formulae about the
thermodynamics of an open system subject to uneven energy transfers,
since there are plenty of books out there that already do it.

[Dallas]

On Wed, 08 Aug 2007 06:57:54 -0700, Doug Semler wrote:

> But that doesn't change the fact
> that I don't think anyone in *this* group wants me to dust off the 'ol
> applied calculus book and start quoting dry formulae about the
> thermodynamics of an open system subject to uneven energy transfers

Aw... come on... please?

:-)

Thanks to all the respondents

I've actually read everything here and at RAS, but I can see it will take
more than one read to wade through all this.

If we're taking a vote, it looks like the grossly oversimplified answer
winning answer is: b) False

--
Dallas

[d&tm[_2_]]

"Clark" > wrote in message
...
> A minor nitpick on a previous post from Jim:

snip

> It was stated that n/V was equal to density. While the stated equality
> isn't,
> the jist of what was said was ok. The equality would be
>
> (Mass/Molecular Weight)/V = density
>
> Since the molecular weight of air is a constant it can be combined with
> the
> other constant in the ideal gas law when used for atmospheric calcs.
>

a minor nitpick on your minor nitpick is that the molecular weight of air is
not constant. It is affected by humidity.
density is better expressed as d = PM/RT
where d will be denity in units of mass/volume
(M is molecular weight), R the gas constant ,P and T press and Temp . For
us SI people R=8.31, P in Pa and T in Kelvin gives density in kg /m3 (
using M in kg/mole)Terry

[Jim Logajan]

Clark > wrote:
> Nit picking my nitpic? Hmmm, is anyone keeping score? :-) You are
> correct of course but maybe we can take a closer look.
> Lets-see-here-now, we have density in moles/L^3. Is that a nuclear
> density?

The way I see it, if the units of the denominator is length cubed, it's a
density of some sort! That's all a man can ask for - and expect - some
days! ;-)

> Whatever it is it must be a Chem E thing... I've seen flow
> rates in moles/hour, but this is new. Ya never can trust those Chem E
> types anyways.

Add physics grads to the types you can't trust. ;-) You do realize that
physicists consider 1, pi, and 4.9 to all equal 1, while 11, 33, and 49 are
all equal to 10? They love to round to the nearest power of 10, so they can
reduce all multiplcations and divisions to additions and subtractions of
powers of ten. We never learned math beyond adding and subtracting. :-)

> I figured you chose to skip a few steps to get to the answer sooner
> and avoid losing the audience. It's a tough call on the minor
> technical details. I probably err in the other direction too often. I
> figured this group has some folks that wouldn't mind seeing an
> expanded explanation in a separate post.

I was sure people's eyes had glazed over before they finished reading.
What's a few lazy mistakes among friends, eh?

While I'd love to expand on my explanation, as I write this I believe a new
episode of Mythbusters comes on in ~15 minutes and I know I wont be able to
compose an accurate post in that amount of time. And my short attention
span means I'll never get round to it later. (Did I mention I did better in
QM than statistical and thermo physics and that I'm really rusty in both
now? Or as Scotty once said: "I canna change the laws of physics Captain! I
need 30 minutes!")

[d&tm[_2_]]

"Clark" > wrote in message
...
> "d&tm" > wrote in
> :
>
>>
>> "Clark" > wrote in message
>> ...
>>> A minor nitpick on a previous post from Jim:
>>
>> snip
>>
>>> It was stated that n/V was equal to density. While the stated equality
>>> isn't,
>>> the jist of what was said was ok. The equality would be
>>>
>>> (Mass/Molecular Weight)/V = density
>>>
>>> Since the molecular weight of air is a constant it can be combined with
>>> the
>>> other constant in the ideal gas law when used for atmospheric calcs.
>>>
>>
>> a minor nitpick on your minor nitpick is that the molecular weight of
>> air is not constant. It is affected by humidity.
>> density is better expressed as d = PM/RT
>> where d will be denity in units of mass/volume
>> (M is molecular weight), R the gas constant ,P and T press and Temp .
>> For us SI people R=8.31, P in Pa and T in Kelvin gives density in kg
>> /m3 ( using M in kg/mole)Terry
>>
>>
> I use standard dry lab air in all my calculations. :-)
>
> I vote we call this one a sub minor nit pick on my minor nitpick. Just
> what
> is the likely range of average molecular weights? Water vapor content
> varies from what? pretty darn dry to about 4%? And keep those SI units on
> your side of the border, or pond, or whatever socio-economic,
> nation-state,
> or geographic boundry that thankfully separates us.
>
4% water could make the density altitude ( the important thing to us pilots)
10 to 20% higher than if the air was dry. Thats not insignificant to me.
But then I know a lot of enjuneers who would be happy with that. ( mostly
the ones who cant spell enjuneer) . and wots wrong with SI units Clark ?
God gave us 10 fingers for a reason.( at least we have 10 fingers on this
side of the pond as you call it, not so sure about you murkins :<)
terry

[Denny]

I seem to have come late to this scrum...

Anyway, the last poster missed a thing or three... Like Euler, like
Boyles law, like the equation for lift, like Bernouille, etc..

The biggest thing he missed is if you heat a gas and keep its pressure
the same then it has to expand (That old Boyle guy again)
Expanded gas means the molecules are farther apart... Molecules being
farther apart means less density, less density means less lift...

Ya know in the ancient times of BI (before internet) you had to truck
down to the library and dig through piles of books to find this
information... Today, in 0.002 seconds your web browser will pull up
the articles and equations on this subject... It is highly recommended
by me...

denny

[Tom L.]

The statement (taking into account the later added context, i.e.
Jeppesen's text) is technically true, although not very useful in the
aviation environment.

As Jim L. pointed out, the presure is proportional to gas density and
temperature. That's all there is to it (the pressure is NOT the weight
of the the column of air). So when Jeppesen says "all else being
equal" while analyzing temperature and pressure, that "else" has to be
density. If density stays constant, higher temperature will result in
higher pressure.

But airplanes don't fly in an enclosed container where one can keep
the density constant. For example: higher temperature (all else being
equal :-) ) will tend to deacrease the density. And then there are
winds, ...

So the statement (although technically correct) is at best pointless,
and at worst (dangerously ?) misleading.

- Tom

On Mon, 6 Aug 2007 12:05:17 -0500, Dallas
> wrote:

>Brought over from RAS:
>
>
>Assuming that all other variables remain constant:
>
>An increase in temperature will result in a higher atmospheric pressure - a
>higher temperature speeds up the movement of the air molecules, thereby
>raising the pressure they exert on the surrounding atmosphere.
>
>A) True
>B) False

[Dallas]

On Mon, 6 Aug 2007 12:05:17 -0500, Dallas wrote:

> An increase in temperature will result in a higher atmospheric pressure - a
> higher temperature speeds up the movement of the air molecules, thereby
> raising the pressure they exert on the surrounding atmosphere.

Most respondents discounted this statement as incorrect or a flawed
problem. And yet it appears very simply stated in this form as a test
question on the private pilot written:

How do variations in temperature affect the altimeter?

A. Higher temperatures expand the pressure levels and the indicated
altitude is higher than true altitude.
B. Lower temperatures lower the pressure levels and the indicated altitude
is lower than true altitude.
C. Pressure levels are raised on warm days and the indicated altitude is
lower than true altitude.


Answer (C) is Correct - On warm days, the atmospheric pressure levels are
higher than on cold days. Your altimeter will indicate a lower than true
altitude. Remember, "Low to high, clear the sky."

:-/

--
Dallas

[Tom L.]

On Mon, 13 Aug 2007 17:26:59 -0500, Dallas
> wrote:

>On Mon, 6 Aug 2007 12:05:17 -0500, Dallas wrote:
>
>> An increase in temperature will result in a higher atmospheric pressure - a
>> higher temperature speeds up the movement of the air molecules, thereby
>> raising the pressure they exert on the surrounding atmosphere.
>
>Most respondents discounted this statement as incorrect or a flawed
>problem. And yet it appears very simply stated in this form as a test
>question on the private pilot written:
>
>How do variations in temperature affect the altimeter?
>
>A. Higher temperatures expand the pressure levels and the indicated
>altitude is higher than true altitude.
>B. Lower temperatures lower the pressure levels and the indicated altitude
>is lower than true altitude.
>C. Pressure levels are raised on warm days and the indicated altitude is
>lower than true altitude.
>
>
>Answer (C) is Correct - On warm days, the atmospheric pressure levels are
>higher than on cold days. Your altimeter will indicate a lower than true
>altitude. Remember, "Low to high, clear the sky."
>
>:-/

"Everything else being equal". But it rearely is.

Consider the following example. The numbers are temperatures and
pressures as reported by six airports in San Francisco Bay area about
an hour ago. Flying from SFO to RHV one flies over (or close to) SQL,
PAO, NUQ, and SJC, in that order. The total distance from SFO to RHV
is a little over 30 nm.
Check the numbers:

SFO 21 30.01
SQL 25 29.99
PAO 24 29.99
NUQ 23 30.00
SJC 26 29.98
RHV 28 29.99

The temperature *increases* significantly, white the pressure
*decreases" (or stays the same if we assume that 0.1% variation is
within measurement tolerances).

We can also look at temperatures and pressures in one place changing
over time, for example at SFO over the last eight hours:

15Z 14 30.07
16Z 16 30.08
17Z 17 30.08
18Z 18 30.07
19Z 19 30.06
20Z 21 30.05
21Z 21 30.03
22Z 21 30.02
23Z 21 30.01

The numbers wouldn't make FAA question designers proud.

- Tom