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

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Air is colorless despite nitrogen and carbon dioxide being white gases because the visible white fumes observed during the evaporation of liquid nitrogen or dry ice are actually due to the condensation of water vapor, not the gases themselves. The discussion includes various scientific teasers aimed at college-level understanding, including calculations related to planetary formation and the behavior of sound in a vacuum. Participants are encouraged to solve these teasers while adhering to specific rules about attempts and timing. The conversation also touches on misconceptions about sound propagation in a vacuum and gravitational potential energy in relation to Earth's shape. Overall, the thread promotes engagement with complex scientific concepts through problem-solving.
  • #31
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.
 
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  • #32
Number 5. Remember, vectors have two important characteristics. They have a magnitude and a direction.
 
  • #33
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.

Gokul43201 said:
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.
 
  • #34
Gokul43201 said:
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. :eek:

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. :wink:

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.
 
  • #35
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. :wink:
 
  • #36
Gokul43201 said:
This reason comes from the acoustic mismatch of air and glass.
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huh, interesting point! that's what i like physics for :)
thanks for clarification!
knowing that You are busy, it sounds slightly cruel but anyway: "looking forward to Part 2" :))