Why is air colorless even though nitrogen and carbon dioxide are white gases?

[Gokul43201]

THE RULES :

1. Each person may attempt only one teaser, total, unless I announce otherwise.

2. If teasers remain unsolved after a significant length of time ( a couple of days), I will announce a 'free for all', at which time anyone (including those that have already answered) may attempt any number of the remaining questions.

3. Anyone may attempt the bonus question, even if they have already tried another question.

Note :

1. Some of these teasers require back-of-envelope type calculations, or 'order of magnitude' estimates.

2. All teasers are roughly aimed at the college/grad level. Some are accessible to high schoolers.

3. Please read the rules above and follow them.



THE TEASERS :


1. When liquid nitrogen or dry ice evaporates/sublimes, you can see thick, white fumes. If nitrogen and carbon dioxide are really white gases, why is air colorless ?



2. A planet forms from a large, cold cloud of dust. Neglecting radiation, find the radius of the planet (order of mag.) if its temperature keeps the planetary material at about its melting point.

Show working/explanation.



3. Experimental particle physicists have elaborate experiments set up (some, deep down in modified mine shafts) to observe proton decay. Derive a lower limit (very rough number) on the lifetime of the proton on the based on the fact that you are alive.

Assume that a dosage of about 600 rad is lethal. Also assume that a proton always decays into a positron and a neutral pion. Use Google for decay info or for conversions.



4. Sound can not propagate through vacuum.

In a popular demonstration of this, an electric bell is set up inside a bell-jar, which is then evacuated using a vacuum pump. When the air pressure drops below about 1cm of Hg, the bell is no longer audible. Voila !

Show why this demonstration is spurious, and, if you can, provide the correct interpretation.


5. BONUS QUESTION :

Resolve this paradox.

The Earth is not a sphere, but is, approximately, an oblate spheroid, with its equatorial diameter a about 13 miles (21 km) greater than its polar diameter b. Modeling this in 2 dimensions, as an ellipse, you have $x = a~cos \theta, ~~y=b~sin \theta$. Writing $a-b=\delta$ and neglecting squares of $\delta /a$, the equation yields $r = \sqrt{x^2 + y^2} = a - \delta sin^2 \theta$, where $\theta$ is the latitude. This gives $r_{30} - r_{50} = 4.6~$mi (about 7.4 km). So a point on the sea level at a latitude of 30N is about 4.6 miles higher than a similar point at 50N.

Now the Mississippi River flows from roughly 50N to about 30N, where it drains into the Gulf of Mexico. The altitude at the source is about 1.5 miles above mean sea-level, and the mouth is obviously at sea-level. Now, what this means is that the river has effectively climbed a height of about 3 miles in its journey to the sea.

But we all know that water can not flow upwards against gravity, from a smaller height to a greater one. So the Mississippi is a figment of our imagination.

That's all for now...more coming up after this.

 

[DarkEternal]

i am going to take a quick guess at the bonus question. would it have anything to do with the centrifugal fictional force caused by the Earth's rotation?

 

[Gokul43201]

I won't say whether or not the bonus question has been answered correctly, till the end (say in a few days or so). I want everyone to give it a shot.

 

[NoTime]

1-> Condensed water vapor

5-> Um. You pretty much give the answer in the question.
>So a point on the sea level at a latitude of 30N is about 4.6 miles higher than a similar point at 50N.
Sea level defines the gravitation equipotential surface.

 

[Gokul43201]

No Time, you're right on Q1. But now, you may not attempt any more until I throw it open to all.

No comment on Q5.

No takers for 2, 3, 4 ? Fraidy cats, eh ?

 

[hemmul]

N4:

-- [select]
hmm...
a bell can't hang in the air freely. it needs some support (a basement, a rope) e.i. something connects it to a surface of the bell-jar. The sound propagates directly through this material, and reaches us (of course it is slightly disrtorted). Now the only way out i see is to "hang" the bell inside the jar with a magnet or something, avoidint the direct contact ;)
-- [deselect]

 

[hemmul]

N5:

actually the heights that differ are calculated respectively to the geometrical center of the ellipse (and there is no surprise in that) If that was so for a sphere, it would be strange - because the potential phi~1/r. Now what about ellipse?
Consider an ellipse with axes a & b. now let's expand/shrink the coordinate system (e.i. b/a-factor scaling transformation) via OX axis so, that the ellipse transforms into a sphere of a radius b. everywhere on the surface of a sphere the potential~1/b. now "i don't think" that the fact of the surface "equipotentiality" will change after a backward transformation to the ellipse ("i don' think" - because i believe it can be prooved, making certain substitutions in the potential integral for an ellipse, but i have no back-of-envelope nearby right away :rofl:)
So, the potential everywhere on the surface is same (no wonder - the ocean water chooses least-potential states), and there is no potential-energy losses during Misissipi's voyage! However, the height relatively to the centre of the Earth does change.

So the Mississippi is a figment of our imagination.

maybe i have actually never seen it...

 

[Gokul43201]

hemmul,

You've made an interesting point for Q4, but that would only be a reason for you to hear the bell ringing, not the opposite.

 

[NateTG]

THE RULES :

5. BONUS QUESTION :

Resolve this paradox.
...

The paradox makes the tacit assumption that surfaces of gravitational equipotential on the surface of the Earth are spherical while explictly modeling the Earth as an elipsoid, but elipsoids generally do not have spherical surfaces of equal gravitational potential.

Moreover, there is an additional force that is not being accounted for: the Earth is a rotating reference frame, and the Mississippi is traveling with the action of the centrifugal force.

 

[Bartholomew]

4.)I'm going to go out on a limb and say that the demonstration is valid. I was thinking that the speed of sound might have been decreased by the lowered air pressure and that this speed change could cause the sound to be reflected internally within the bell but wikipedia tells me that air pressure has no effect on the speed of sound. With that idea discarded, I can't think of any reason why there would be silence except that as the air becomes vacuumlike there are just fewer particles to transmit the sound.

 

[DaveC426913]

4. Well, I don't know about 'spurious', it certainly demonstrates the point in layman's terms (you said "...popular demonstration..."), but it is not rigorous science, no.

For one, it is not in vacuum, it is only in partial vacuum. You cannot make more than a good partial vacuum on Earth using pumps.

 

[DaveC426913]

4. Waitaminnit, when offered the "sound wuld be transmitted through the support" explanation, you responded with "...only be a reason for you to hear the bell ringing, not the opposite..."

So, you're looking for an answer that shows that:
- you would not be able to hear the bell ringing, AND
- that is not proof that sound doesn't travel through a vacuum
?

Which means you propose that the bell is not heard for some other reason.

Is this correct?

 

[Gokul43201]

Yes, that's right.

Bart came close, but I don't have time to explain now. Got a flight to catch.

 

[Rogerio]

4.
In some situations, the sound transmitted through the air could be canceled by the sound transmitted through the solid material of the bell-jar. This effect depends on some parameters (phase and transmission coeficients) but it could be possible.

 

[NoTime]

4.[]

Well, since Gokul is flying off somewhere I guess he won't mind me saying that's not it for him.

 

[Rogerio]

Hi Gokul,
the answers, please!

 

[Gokul43201]

Oops...sorry. Answers tonight.

 

[NoTime]

Oops...sorry. Answers tonight.

Does that mean I can answer them now

4) Bell frequency change.

3) Opinions to the contrary, protons do not decay.

I think the first time I saw this idea I think the estimate was 10^17 years.
Not sure what its up to now, ?10^133 or something?
Perhaps we would have acquired some of the properties of radiodurans.

2)This one I originally thought there was not enough info for.
Now I'm thinking that you might have the idea of using something like the gas law, then cross this with the mass required to provide the appropriate pressure.

 

[Gokul43201]

1. When liquid nitrogen or dry ice evaporates/sublimes, you can see thick, white fumes. If nitrogen and carbon dioxide are really white gases, why is air colorless ?

This one has been answered correctly by NoTime.

The white fumes are from the condensation of water vapor due to the local cooling of the air below its dew point.

That was just to get you warmed up. This is the only question among 1,2,3,4 that has satisfactorily been answered.

Since I completely forgot about this thread and I promised a "free for all" after a certain period of inectivity, this is that free-for-all. Anyone may attempt any number of questions among 2, 3, 4, 5.

I'll acknowledge correct solutions, or else post my solutions around this time tomorrow. Remind me, if I forget.

 

[Gokul43201]

2. A planet forms from a large, cold cloud of dust. Neglecting radiation, find the radius of the planet (order of mag.) if its temperature keeps the planetary material at about its melting point.

Show working/explanation.

The gravitational potential energy lost by the dust cloud upon coming together must go into thermal energy gained by the planet thus formed.

The gravitational PE of a sphere of mass M, radius R, is $\frac{3GM^2}{5R}$. Since the initial cloud was considered large, its PE ~ 0.

If you didn't know this you could have either derived it or gone with the rough value $\frac{GM^2}{R}$.

$$\frac{3GM^2}{5R} = MCT_m~, ~~M = \frac{4\pi R^3 \rho}{3}$$

$$=> R^2 = \frac{5CT_m}{4\pi G \rho }$$

Now plugging in typical (rough) values

$$C = 10^3~ J/K \cdot kg~,~~ T_m = 2*10^3~ K~, ~~\rho = 5*10^3~ kg/m^3$$

Gives : $$R = (about)~2*10^6 ~m~or~2000~km$$

Notice that this is smaller than the radii of most planets in the solar systems (Radius of Earth ~ 6000 km), suggesting that they were all mostly molten when they formed, but have eventually cooled off through radiation. Indeed the inner core of the Earth is still molten.

 

[gerben]

I am very curious of the answer to question 4

 

[Gokul43201]

I'll get to it by tonight or tomorrow.

 

[K.J.Healey]

4)
Is it spurious because it states that the bell is not "audible"? What is audible? Is it what the human ear can hear or is it ANY change in pressures that defines sound. Similar to asking if all light is visible. Therefor, being that the container could never be an absolute vacuum, and particles exist inside it, and will then be succeptable to pressure changes by local volume changes(the movment of the bell/buzzer), so there WILL be some transfer of energy to the "air" particles in the pseudo-vacuum. That is sound (if that is how we choose to define it).


Theres my answer, I hope I am right.

 

[K.J.Healey]

Have you had time to type up the answers yet?

 

[NoTime]

The gravitational potential energy lost by the dust cloud upon coming together must go into thermal energy gained by the planet thus formed.

The gravitational PE of a sphere of mass M, radius R, is $\frac{3GM^2}{5R}$. Since the initial cloud was considered large, its PE ~ 0.

Well, I couldn't argue with wine bottle math
However, this is close enough to what I had in mind that I think I understand what you did.

So, why would the PE of the cloud ~ 0?
I would think that just the opposite would be true.
A dust particle gains potential energy with height, or distance from the planet center in this case.

If I read what you wrote correctly, you would have to turn your planet into a point in order to extract the potential energy you are equating to temperature.

What am I missing here.

 

[Gokul43201]

The initial dust cloud is said to be "large" compared to the planet. So, the assumption is that you start with a (nearly infinitely) dispersed cloud. This means that the distance between particles in the cloud is large enough that the potential energy is negligible (PE ~1/r).

This is just the many particle analogue of having 2 particles initally far apart. The potential energy of the pair is 0 (-Gm1m2/d ->0, for large d) to start but some negative number as they get closer. Of course, this loss of potential energy is accounted for by the gain in KE...till eventually the 2 particles are at the point of collision. Now, all the lost potential energy has been converted to KE. But now assume that the particles collide and stick. The KE goes to 0, and all the energy goes into heat. So basically, the PE difference (to a rough approximation, at best) will give you the heat produced.

Okay, let me write up the next one now.

 

[Gokul43201]

3. Experimental particle physicists have elaborate experiments set up (some, deep down in modified mine shafts) to observe proton decay. Derive a lower limit (very rough number) on the lifetime of the proton on the based on the fact that you are alive.

Assume that a dosage of about 600 rad is lethal. Also assume that a proton always decays into a positron and a neutral pion. Use Google for decay info or for conversions.

The decay products are given to be $e^+$ and $\pi ^0$. The provided link shows that
$$\pi ^0 \rightarrow e^+ + e^- + \gamma \rightarrow 2\gamma$$
The positron ($e^+$) will clearly be annihilated by the nearest available electron, releasing yet more energy. So, we can assume that nearly the entire mass of the proton goes into energy.

Thus a mass, m, of protons, will produce energy $E/m = c^2 \approx 10^{17}~J/kg$

Since the decay, and hence the energy produced goes like $$E_0e^{-t/\tau }$$
the initial dose rate goes like $$E_0/\tau~ J/kg \cdot yr$$

1 J/kg = 100 rad. So, if you're about 20 years old, and alive

$$\frac{20*100*10^{17}}{\tau} < 600$$

Which gives, $$\tau > 3*10^{17}~ yrs$$

A more careful analysis will consider that you didn't weigh m kgs for all of the 20 years as well as losses to lower probability decay routes (~10% usually does not get converted to energy) and a host of other lower order effects.

To raise the lower limit on the lifetime, you have to wait longer and have more mass. This is what proton decay experimentalists do. They get themselves a giant tank of water...and they wait !

 

[NoTime]

Thanks for the clarification

 

[hemmul]

N4 & N5 pleeeaze ;)

 

[Gokul43201]

Yes, I'm sorry. Will do it today or tomorrow. Been busy.

 

[Pseudopod]

5 seems pretty straight forward. Centrifugal force shouldn't have anything to do with it because all these measurements are taken with respect to sea level. The ocean is one big pool of water and it will naturally form a surface of equipotential gravity. This means the oceans will be bulged at the equator (because gravity is less) so the height difference is already accounted for by the water. Another way of thinking about it is this: Imagine a blob of water a few miles above sea level at 50N. This would certainly sink into the ocean and disappear. If it was true that water at 50N had to climb a few miles against gravity to get to water at 30N then we'd have a huge flow of water from 30N flowing with gravity to 50N and it would soon equalize anyway. This latter point might be circular reasoning working backwards from the original question, but the key to it all is you are measuring height relative to the local sea level which is the same as measuring it with respect to equal gravity.

 

[BobG]

Number 5. Remember, vectors have two important characteristics. They have a magnitude and a direction.

 

[Gokul43201]

This has been long overdue. My apologies to all, for keeping you waiting...I've been a little tied up with work...and it takes more than a couple of minutes to do the calculations and write it all out.

4. Sound can not propagate through vacuum.

In a popular demonstration of this, an electric bell is set up inside a bell-jar, which is then evacuated using a vacuum pump. When the air pressure drops below about 1cm of Hg, the bell is no longer audible. Voila !

Show why this demonstration is spurious, and, if you can, provide the correct interpretation.

No one really attempted the first part of this problem, though some came pretty close to the second.

To show that this demonstration is spurious (ie : that the reason we do not hear the bell is not because sound can not propagate through vacuum) :

Sound will propagate normally, even at reduced pressures, as long as the mean free path is small compared to the wavelength. At 1 cm of Hg, the mean free path $l \approx 10^{-6} ~m$. But the typical wavelengths of audible sound are of order several centimeters. So, clearly the pressure is not low enough to prevent the propagation of sound. There must be some other reason we do not hear the bell.

This reason comes from the acoustic mismatch of air and glass. When a plane wave is incident normally on the plane interface between two media A and B, a fraction R, of the energy is reflected, given by $$R = \frac{(Z_A - Z_B)^2}{(Z_A + Z_B)^2}$$
Z being the acoustic impedance : $$Z = \sqrt{\rho E}$$
where $\rho$ is the density and E is the elastic modulus. The fraction transmitted is 1-R. When $Z_A$ and $Z_B$ are very different, this number is approximately $4Z_A/Z_B$, where $Z_A << Z_B$. (These last couple of equations will look very familiar to anyone who's solved the Schrodinger Equation for a particle in finite rectangular potentials.) For a gas at pressure p, the bulk modulus is given by $\gamma p$, so the impedance goes like $\sqrt {\rho p}$ which in turn is just proportional to p. Thus the fraction of sound energy transmitted will fall as the pressure is reduced.

While this gives some insight into the physical mechanism involved, it is hardly a complete explanation for the problem involved. The problem has analytical solutions for plane surfaces and plane waves, but in this case, neither is plane. Nevertheless, the general result that the impedance mismatch gets worse as the pressure is reduced remains valid.

An alternative approach is to model the system as two masses (the bell and the glass) connected by a weak spring (the air). The stiffness of the spring corresponds to the gas pressure. It can be shown that the amplitude of the vibration of the jar will also be proportional to p, and hence, the intensity of sound transmitted goes away like $p^2$.

 

[Gokul43201]

5. BONUS QUESTION :

Resolve this paradox.

The Earth is not a sphere, but is, approximately, an oblate spheroid, with its equatorial diameter a about 13 miles (21 km) greater than its polar diameter b. Modeling this in 2 dimensions, as an ellipse, you have $x = a~cos \theta, ~~y=b~sin \theta$. Writing $a-b=\delta$ and neglecting squares of $\delta /a$, the equation yields $r = \sqrt{x^2 + y^2} = a - \delta sin^2 \theta$, where $\theta$ is the latitude. This gives $r_{30} - r_{50} = 4.6~$mi (about 7.4 km). So a point on the sea level at a latitude of 30N is about 4.6 miles higher than a similar point at 50N.

Now the Mississippi River flows from roughly 50N to about 30N, where it drains into the Gulf of Mexico. The altitude at the source is about 1.5 miles above mean sea-level, and the mouth is obviously at sea-level. Now, what this means is that the river has effectively climbed a height of about 3 miles in its journey to the sea.

But we all know that water can not flow upwards against gravity, from a smaller height to a greater one. So the Mississippi is a figment of our imagination.

Easy one. Several folks got this right.

Equipotentials are not spherical shells. And while this is partly due directly to centrifugal effects, it is more a result of the non-sphericality of Earth (which in turn is largely due to centrifugal effects). One equipotential is clearly sea level itself, and since the Mississippi flows from a greater height from sea level, to a lower one, it does not increase its potential energy by doing this.

 

[Gokul43201]

I'm not sure I understand BobG's (almost cryptic) clue. Perhaps he's suggesting that the direction of the effective field is not truly radial, but in fact has a slight tilt towards the equator (since the centrifugal vector is parallel to the equatorial plane). This helps things in the northern hemisphere move southwards and things in the southern hemisphere move northwards...but is really only a small effect. The Nile, Ob, Lena, Yenisey, Weser, Elbe, and Parana have had no difficulty flowing away from the equator.