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Problems about sound, light, and earthquakes.

1. Lightning and earthquakes

ligtning
Problem 1

We see a flash of lightning. Three seconds later, we hear thunder. How far away was the lightning bolt?

Solution

Sound travels at about 340 meters per second in air under "standard conditions" (temperature, pressure, humidity). Light travels very fast, about 300,000 km/sec in the vacuum. For all practical purposes, there is no delay between the lightning strike and the viewer’s perception of the lightning flash some kilometers away. Thus the interval of time between seeing the flash and hearing the thunderclap is, for all practical purposes, the same as the amount of time it took for sound to travel from the site of the lighting to you. Since \[ distance = speed \times elapsed\ time, \] we have \[ distance = 340\times 3 = 1020\ meters \]

Problem 2

There is a flash of light. Just 2.1 seconds later, thunder is heard. How far away was the lightning strike?

Earthquakes

When an rock breaks underground because it can no longer support the stress, energy is released, sending waves of various kinds through the earth. One kind of wave is the P-wave — the primary wave. It is a wave of longitudinal compression and decompression, just like sound waves in air. P-waves travel at about 8 km/sec. The actual speed depends on the type of rock. S-waves are transverse waves (like electromagnetic waves). These are secondary waves; their propagates speed is about 3.45 km/sec. Again, the actual speed depends on the kind of rock.

Scientists have set up measuring instruments all over the world to detect S and P waves. These instruments — seismometers — record the time at which the S and P waves are detected.

Problem 3

(a) How long does it take for a P-wave to travel 1000 km? (b) How long does it take an S-wave to travel the same distance?

Determining how far away the earthquake was

The interval between the arrival of the two waves can be used to determine how far away the earthquake is. Let’s work out one such problem. The P wave of an earthquake is detected at 9:10:07 am. The S wave is detected at 9:13:27 am — exactly 200 seconds later. We know that \[ d = 8t_P \\ d = 3.45t_S \] where \(t_P\) is the travel time for P waves and \(t_S\) is the travel time for S waves.

The problem is that we don’t know \(t_P\) and \(t_S\). We only know the difference of these quantities, \(t_S - t_P\). The actual times cannot be measured, but their difference can be. It is 200 seconds. The difference is an observable quantity.

The good news is that the equations above can be manipulated to give a new equation relating the distance to the difference of the times. This is what we need. To get the new equation, multiply the first equation by 3.45 and the second by 8 to obtain \[ 3.45d = 27.6t_P \\ 8d = 27.6t_S \] Subtract the first equation from the second to get \[ 4.55d = 27.6(t_S - t_P) \] Solve for \(d\): \[ d = 6.07(t_S - t_P) \] We conclude that the earthquake occurred 1214 km away.

Problem 4

The P-wave from an earthquake arrived at 2:34:22 pm. The S-wave arrived at 2:40:51 pm. How far away was the earthquake?

Problem 5

(a) A seismic station located in Los Angeles records the time of arrival of S and P waves and determines that an earthquake has occurred 2000 km away. What can we say about the location of the earthquake? (b) A seismic station in Seattle, 1545 km away, detects the same earthquake and determines that it occurred 1200 km away. Is this enough information to determine the location of the earthquake? Think about how we determine the location of an earthquake.

Tsunamis

The speed of a tsunami wave depends on depth and wavelength. Deep water waves move faster than shallow water waves. Below is a table taken from hyperphysics

Depth (meters)

Velocity (km/h)

Wavelength (km)

7000

943

282

4000

713

213

2000

504

151

200

159

48

50

79

23

10

36

10.6

Problem 6

An earthquake is detected 1000 km from the coast of Japan. How much time to prepare do they have before the tsunami arrives. (Estimate)

To memorize for future use
  1. The velocity of sound in air under standard conditions.

  2. The velocity of P and S waves. (Note that a P wave is sound wave, but where the medium is rock, not air.

How long does light take to travel form the sun to earth? How long from one end of a ping pong table to the other? How does light bounce of a mirror? How do ping pong balls bounce?

2. Light and photons

Introduction

Newton thought of light as a particle (now called a photon). Huygens thought of light as a wave. Both are right — a fact which we can understand using quantum mechanics. Thus it makes sense to say

  • A photon of light travels through a vacuum at 300,000 km/s.

  • A light wave travels through a vacuum at 300,000 km/s.

Problem 7

How long does it take for a photon to travel from the sun to the earth?

Problem 8

How long does it take for a photon to travel the length of a ping pong table?

Problem 9

How long does it take sound to travel the length of a ping pong table?

Problem 10  — Experiment

How long does it take for a ping pong ball to travel from one end of a ping pong table to the other? You can try to devise an experiment to find this quantity. You can also make an arm chair estimate.

Some facts
  1. Velocity of light: 300,000 km/sec

  2. Distance of sun to earth: 150,000,000 km

  3. Length of ping pong table: measure!

To memorize for future use
  1. The velocity of light

  2. The distance from the sun to the earth

How does light bounce off a mirror? How do ping pong balls bounce?

3. Reflection: photons and ping pong balls

mirror

Imagine a photon approaching a mirror \(M\) as in the figure. The line \(N\) is perpendicular to \(M\). It is called the normal line. When the photon reaches the mirror, it bounce up and to the right. The photon has been reflected.

Note the angles \(\alpha\) and \(\beta\) in the figure. These are the angle of incidence and the angle of reflection. The law governing reflection of photons is that \(\alpha = \beta\): the angle of reflection equals the angle of incidence.

Problem 1.
A photon approaches a mirror. The angle of incidence is \(20^\circ\). Make a drawing that shows the incoming ray, the normal, and the outgoing ray. On the same drawing, show rays with an angle of incidence of \(45^\circ\) and of \(70^\circ\).
Problem 2.
Imagine that the incoming particle is a ping pong ball, and that the mirror is a ping pong table. Does the same law of reflection hold? Examine the same question for billiards (pool). What about other particles?
corner
Problem 3.
In the figure on the left, you see two mirrors attached to each other at right angles. Make a copy of this drawing and complete it, showing whatever rays (photon paths) are needed. What can you conclude about the incoming and outgoing rays. (By outgoing rays, we mean the one that will "leave the figure". Finally, imagine three mirrors, each at right angles to one another, joined like the walls that meet in the corner of the room. What happens when a light ray strikes this "corner" mirror.
mirror angle small
Figure 1. Mirror A
Problem 4.
In the figure on the left photons strike a mirror, traveling from right to left. The incoming rays indicate photon movement. In this case, the angle of incidence \(\theta\) is small. + Make a copy of this drawing, measure the angle of incidence, and draw the outgoing (reflected) rays. Measure the spacing of the incoming rays and of the outgoing rays. Compute the ratio \[ \rho_A = \frac{\text{outgoing spacing}}{\text{incoming spacing}} \]
mirror angle large
Figure 2. mirror B
Problem 5.
Carry out the same steps for mirror B that you did for mirror A. Take care to make the spacing between the incoming rays the same as the in the first figure. Let the ratio for the second mirror be \(\rho_B\). How does the ratio \(\rho\) affect the intensity of the reflected light?
mirror ray spacing
Figure 3. Ray spacing
Problem 6.
Consider once again mirrors A and B. Let \(s_\text{in}\) be the spacing between the indicated rays. Let \(s_A\) and \(s_B\) be the spacing between "points of impact" of adjacent rays, as indicated in the figure. Measure these quantities in your drawings. Imagine now that the mirror is a black surface and that the two surfaces A and B are illuminated equally. Which one will absorb heat more rapidly? Can you make a numerical estimate of the rate of energy transfer (light to heat) for the two surfaces?
A better way to work with very large and very small numbers.

4. Scientific notation

Scientists have a clever way of writing very large and very small numbers. Instead of writing 7,000,000,000, they write \(7\times 10^9\). In scientific notation, there are two parts, the coefficient and the power of ten. The exponent in the power of then tells you how many zeroes to put after the coefficient.

Problem 1. (a) Write 400,000,000,000,000,000 in scientific notation. (b) Write \(8\times 10^{19}\) in ordinary notation.

In scientific notation, the number 1,300,000,000,000 is written \(1.3\times 10^{12}\). This time the coefficient is not a whole number. But it is a number bigger than zero and less than ten. We usually put the coefficient in this form. The exponent then tells you how many places to the right the decimal point should be moved to get the usual representation of a number.

Problem 2 (a) Write 76,400,000,000 in scientific notation. (b) Write \(3.4\times10^{22}\) in normal notation.

Often times we drop digits that we don’t think are significant. Thus, instead of writing 12,345,678,988 for the number of bacteria in a sample, we write \(1.2\times 10^{10}\).

Example: The light-year

Light travels about 300,000 km per second, or 300,000,000 meters per second. How far does it travel in a year? This distance is by definition the light year. Well, in a year there are \(365\times24\times60\times60 = 31,536,000\) seconds. The distance light travels in a year is then \[ 300,000,000\times31,536,000 = 9,460,800,000,000,000\text{ m} \] It is hard to comprehend such a number, even with the commas. In scientific notation this reads \[ \text{One light year} = 9.5\times 10^{15} \text { m} \] This is better, but better still is to do the computation entirely in scientific notation \[ 3\times10^8\times 3.2\times 10^7= 9.6\times 10^{15} \text { m} \]

Rounding to the first significant place, \[ \text{One light year} = 10^{16} \text { m} \]

Problem 3 — Mass and energy

Einstein discovered the relation \[ E = mc^2 \] that relates the rest mass of an object to the energy that is released if all that mass is converted into energy. (a) Find the energy in Joules released by converting one gram of matter into energy. Work in standard units — the MKS system (meters, kilograms, seconds) throughout. Then your energy will be in Joules. (b) How many liters of gasoline would you have to burn to release an equivalent energy?

Example

The diameter of the earth is 12,742 km. We are going to estimate its volume by thinking of it as huge cube with sides 10,000 km long. (We will do a more accurate computation later). A cube with sides of length \(s\) has volume \(V = s^3\). Since we want the answer in meters, we first write \(s\) in meters:

\[ s = 10^4 \text{ km}\times 10^3 \text{ m per km} = 10^7 \text{ m} \]

Now we compute the volume:

+ \[ V = s^3 = 10^{21} \text{m}^3 \]

Problem 4. The diameter of the moon is 3474.8 km. (a) Use the example as a guide to estimating the volume of the moon. (b) Compute the ratio of the diameter of the earth to the diameter of the moon. (c) Compute the ratio of the volume of the earth to the volume of the moon. (d) How are (b) and (c) related?

Surface area and volume of a sphere

The surface area of a sphere is given by the formula

\[ A = 4\pi r^2 \]

The volume of a sphere is given by

\[ V = \frac{4}{3}\pi r^3 \]

Problem 5. Compute the surface area of the moon and earth in square meters.

Problem 6. Compute the volume of the moon and earth in cubic meters.


Table
Prefix Value

kilo

\(10^3\)

mega

\(10^6\)

giga

\(10^9\)

peta

\(10^{12}\)


Scientific notation handles small quantities using negative exponents.

5. Very small quantities

Scientific notation can also handle very small quantities. As warmup, we have this:

\[0.004 = \frac{4}{1000} = \frac{4}{10^3} = 4\times10^{-3}\]

It is easy to see the rule: If you move the decimal point \(n\) places to the right to get the coefficient, then the exponent is \(-n\).

Example

A water droplet has a diameter of about 1 millimeter, hence a radius of about \(5\times10^{-4}\) meters. Its volume is

\[V = \frac{4}{3}\pi\, (5\times 10^{-4})^3= 5.2\times 10^{-10} \text{m}^3\]

Problem 7. The streptococcus bacterium is nearly spherical in shape and has a diameter of about 1 micron, or \(10^{-6}\) meter. What is the volume of the streptococcus bacterium in cubic meters?

Problem 8. About how many streptococci can you pack in one drop of water?'

Prefix Value

milli

\(10^{-3}\)

micro

\(10^{-6}\)

nano

\(10^{-9}\)

pico

\(10^{-12}\)

Problems about light and sound.

6. Wave length and frequency

Introduction

A wave of any kind has a velocity, a wavelength and a frequency. These quantities are written \(c\), \(\lambda\) (lambda) and \(\nu\) (nu). They are related by the formula

\[c = \lambda\nu\]

Frequency is measured in Herz (cycles per second). If we measure wavelength in meters, then the wave velocity has meters per second as its unit of measure.

According to this relation, the higher the frequency, the lower the wavelength, and vice versa. (Convince yourself of this).

Kinds of waves

There are many kinds of waves. Sound is a wave of compression and decompression in a material medium (air, water, stone, iron, etc). We perceive the frequency of a sound wave as pitch. The higher the pitch, the higher the frequency. The note A of a tuning fork is 440 Herz. Middle C on a piano is about 256 Herz. The low, barely audible hum of household electrical circuits is 60 Herz. Sound waves in dry air at standard pressure and temperature (STP), travel at about 340 meters per second.

Electromagnetic radiation (light, radio waves, x-rays, etc.) is a kind of periodic electrical and a magnetic disturbance. For visible light, frequency (or wavelength) is interpreted as color. Blue light has a wavelength of about 475 nm (nanometers). Red light has a wavelength of about 700 nm.

Sample problem

What is the wavelength of the sound produced by a tuning fork? Substituting the values of frequency (\(\nu = 440\)) and velocity (\(c = 340\)) into the formula, we have

\[340 = \lambda \times 440\]

Solve for \(\lambda\) to get

\[\lambda =340/440 = 0.773\text{ meters}\]
Problem 1

The low-pitched sound made by bad electrical wiring or equipment is 60 Herz. What is the wavelength of this sound?

Problem 2

The highest pitch that a (young) human can hear is about 20,000 Herz. What is the wavelength of this sound?

Problem 3

Imagine a musical instrument made of a pipe of length \(L\) open at both ends — a primitive flute, which you can make with a piece of PVC pipe. The wavelength of the sound produced by this instrument is

\[\lambda = 2L\]

How long should your PVC pipe be to make a sound of A 440? How long should it be if the sound is one octave lower? The frequency of this sound is 220 Herz — exactly one half the original value.

Problem 4

The frequency of the sound used in medical imaging ("ultrasound") has a frequency of 5 to 18 megahertz — 5,000,000 to 18,000,000 Herz. However, the medium in which the sound travels is water, not air. (Why?) Find the speed of sound in water, then find the wavelength of the sound used in medical imaging. Finally, discuss why it is necessary to use such high frequency sounds?

Problem 5

Clamp an iron rod 10 cm long to a vise so that one end is free, one is not. Strike it with a hammer. It makes a sound. The wavelength of the sound produced is given by the formula \(\lambda = 4L\). Given that the speed of sound in iron is about 5000 meters per second, determine the frequency of the sound produced.

Remarks on Musical Instruments

The formula \(\lambda = 2L\) says that the wavelength of the sound produced by our primitive flute is proportional to its length — and that the constant of proportionality is 2. This law is nearly exact for a very thin cylinder — one for which the ratio of length to diameter is large. But in general, size and wavelength are related, and bigger always means longer wavelength.

Remarks on Herz

What the Herz, expressed in fundamental units of measure? Well, by the equation \(c = \lambda\nu\), we have

\[\frac{meters}{second} = (meters)Hertz\]

Solve for \(\nu\) to get

\[ Hertz = \frac{1}{second}\]

Here we have used the fact that units of measure obey the laws of algebra.

The energy of a photon is proportional to its frequency; Planck’s constant.

7. Frequency and energy

Introduction

Remember that light can be thought of either as a particle (a photon) or as a wave. Viewed as a wave, it has velocity, frequency and wavelength.

Now imagine a photon crashing into some matter. Maybe it is a photon bounced off the page of a book you reading. The photon enters your eye, hits the back of the eye (the retina), and releases a tiny bit of energy. This energy excites a rod or cone cell that sends a signal to your brain. Many such photons and many such signals go to make up the image of the printed page you are reading.

The energy carried by a photon is proportional to the frequency of the corresponding wave:

\[ E = h\nu, \]

where

\[ h = 6.62\times 10^{-34}\text{ Joule-sec} \]

is Planck’s constant. Planck’s constant is the fundamental constant of quantum mechanics. It is kind of like the penny in finance. The cost of anything is some integer number of pennies. In Nature, the energy of anything is some multiple of Planck’s constant. Planck referred to this smallest unit of energy as a quantum of energy. The word comes from the Latin quantus, meaning quantity.

Since the wavelength, frequency and wave propagation speed are related, we can also express this in terms of wavelength:

\[ E = \frac{hc}{\lambda} \]

Problem 9

The wavelength of blue light is 475 nanometers. What is its frequency? The wavelength of red light is about 650 nanometers. What is its frequency?

Problem 10

What is the energy carried by a photon of blue light? Of red light? Compute the result in Joules.

Problem 11

When working on the atomic scale, physicists often use the electron volt as the unit of energy. This is the amount of kinetic energy gained by an electron moving through a potential difference of one volt. The electron volt is \(1.6\times10^{-19}\) Joules. Compute the energy in electron volts carried by a photon of blue light. Do the same for red light.

Problem 12 — a story about photoelectrons

A series of experiments Hertz 1887, Hallwachs 1888, J.J. Thompson 1899, Lenard 1902, led to the conclusion that photons can eject electrons from certain metals, e.g, sodium and potassium. Moreover, it was found that the lower the frequency of the light — the longer the wavelength — the less energetic were the ejected electrons. Above a certain wavelength — lets call it the cutoff wavelength — no electrons were ejected. Making the light stronger had no effect. Only changing the color did. The reason for this, it turned out, is that that an electron has to pay an exit fee — a certain amount of energy — to leave the metal plate. If \(E = h\nu\) is smaller than the exit fee, the photon cannot leave. For sodium, the exit fee is 2.28 ev. Find the cutoff wavelength. What color does it correspond to?

Note

It was Einstein who proposed this explanation of the photoelectric effect (1905). It was for this work that he received the Nobel prize.


Type of radiation wavelength

Visible light

400-700 nm

Sun’s ltraviolet light

180-400 nm

Medical X-rays

0.001-10 nm


8. Tables

Table 1. Fundamental Constants
Quantity Value

Avogadro’s number

\(6\times 10^{23}\)

Bohr radius

.0529 nm

Electron volt

\(1.6\times10^{-19}\) Joules

Planck’s constant

\(6.62\times 10^{-34}\text{ Joule-sec}\)

Speed of light

\(3\times10^8\) m/s


Table 2. Energies
Quantity Value

Thermal motion at room temp

0.04 ev

Visible light

1.5 - 3.5 ev

Ionization energy of hydrogen

13.6 ev

Electron striking CRT tube

20 kev

High-energy medical x-rays

200 kev

Beta radiation

0-3 Mev

Gamma radiation

0-3 Mev

Alpha radiation

2-10 Mev

Cosmic rays

1 Mev - 1000 Tev

9. The Bohr Atom

What are atoms made of? What is their structure? We review the "plum pudding" model of Thompson and the "planetary" model of Rutherford, then discuss Bohr’s model, which was the first to incorporate the ideas of quantum mechanics.

9.1. The Back Story - J.J. Thompson

A tentative answer to these questions was formulated by J.J. Thompson in 1897. At that time Thompson was studying the mysterious glow produced when an electric field is applied to an evacuated glass tube — the glow increased with an increase in voltage and a decrease in pressure. Thompson was able to show that the glow was produced by collision of small energetic particles with the atoms of the gas in the tube. He was able to show that the particles carried a negative electric charge, and had mass. While he could not measure the charge and mass separately, he could measure the mass to charge ratio, which he found to be \[ m_e/e = (1.3 \pm 0.02)\times 10^{-11} kg/C, \] where C is the Coulomb: the amount of charge carried by a current of 1 Ampere in one second. Written as a charge-to-mass ratio, Thompson found \[ e/m_e = (0.77 \pm 0.01)\times 10^{11} C/kg. \] The accepted value today is \[ e/m_e = (1.758820088±39) \times 10^{11} C/kg. \] Thus Thompson’s value was accurate to within a factor of two. It took more work to find the values of \(e\) and \(m_e\) separately: \[ e = 1.6021773\;\times\;10^{-19}\;{\rm C} \\ m = 9.109390\;\times\;10^{-31}\;{\rm kg} \] Thompson also measured the charge-to-mass ratio of positively charged \(\ce{H+}\) ions, which found to be about 1000 times smaller than the corresponding ratio for electrons. Faraday’s earlier experiments with electrolysis suggested that the electrical charge of the positive ion and the electron and the ion were the same. Assuming this, it follows that the hydrogen atom is about 1000 times more massive than the electron. In fact, the ration is closer to 1837. While Thompson’s value for the mass ratios was off by a factor of two, the concept of his experiments was correct, as was his theoretical analysis. The main conclusions were therefore that

  • Atoms are made of positive and electrical charges

  • The electrical charges are particles whose mass is about one thousand times less than the mass of a hydrogen atom.

Thompson went on to formulate his "plum pudding" model of the atom — an atom consisted of a kind of blob of positive charge in which the electrons were embedded like raisins in a muffin. The positive charge, equal in magnitude to the sum of the electron charges, would cancel out the negative charges in accord with the observed fact that atoms are normally electrically neutral.

9.2. (D R A F T from here on out)

9.3. Rutherford’s model

10. References

11. Sound waves

11.1. The one-dimensional lattice model

Equation of motion, where \(u_n\) is the position of the \(i\)-th mass, \(M\) is its mass, and \(C\) is the spring constant:

\[ M\frac{d^2 u_n}{dt^2} = -C(2u_n - u_{n+1} - u_{n-1}) \]

Plane wave solution

\[ u_n = Ae^{i(qx_n - \omega t)}, \]

where \(x_n = na\) is the equilibrium position, \(q\) is the wave number, and \(\omega\) is the angular frequency. One finds

\[ \omega = \sqrt{\frac{4C}{M}}\left|\sin \frac{qa}{2}\right| \]

The phase velocity is

\[ v_p = \frac{\omega}{q}, \]

which in the long-wave limit (\(q\) small), is

\[ v_p = q\sqrt{\frac{C}{M}} \]

The group velocity is

\[ v_g = \frac{d\omega}{dq} = a\sqrt{\frac{C}{M}} \cos \frac{qa}{2} \]

In the long-wave limit,

\[ v_g = a\sqrt{\frac{C}{M}} \]

12. Ideal gas law

The ideal gas law in macroscopic units:

\[ PV = nRT \]

  • P = pressure in atmospheres

  • V = volume in liters

  • n = number of moles

  • R = universal gas constant = 0.0821 L·atm/mol·K

  • T = temperature in degrees Kelvin

The ideal gas law in atomic units:

\[ PV = nRT = NKT \]

  • \(N\) = number of molecules

  • \(k\) = Boltzmann constant = \(1.38066 \times 10^{-23}\) J/K = \(8.617385 x 10^{-5}\) eV/K

  • \(k = R/N_A\)

  • \(N_A\) = Avogadro’s number = \(6.0221 \times 10^{23}\) /mol

One mole of an ideal gas at STP occupies 22.4 liters.

12.1. STP

Standard temperature: 0°C = 273.15 K

Standard pressure = 1 atmosphere = 760 mmHg = 101.3 kPa

Standard volume of 1 mole of an ideal gas at STP: 22.4 liters

13. Energy

The unit of energy is the Joule, defined as one Newton-meter.

  • An alkaline AA battery contains about 9360 Joules of energy.

  • Acceleration due to earth’s gravity at sea level is 9.8 m/s. According to Newton’s law, \(F = ma\), a one-kilogram mass exerts a force of 9.8 Newtons. Lifting a one-kilogram mass through a distance of one meter requires 9.8 Joules of energy.

  • An 80 kilogram man runs up a flight of stairs. The distance between the first and second floors is 4 meters. The work done by the man against gravity is 3,136 Joules.

  • A calorie is the amount of thermal energy needed to raise the temperature of water by one degree centigrade. One calorie equals 4.184 Joules. The energy needed to heat one liter of water one degree centigrade is 4,184 Joules.

  • A slice of whole wheat toast has about 128 calories. However, these are the physicists kilocalories. Thus the energy released in metabolizing one slice of whole wheat toast is 128,000 calories — enough to heat one liter of water by 128 degrees C??

  • One gallon of gasoline contains \(1.3\times 10^8\) Joules of energy.

13.1. References

14. Entropy

Quote from [ES]

The second law of thermodynamics says that energy of all kinds in our material world disperses or spreads out if it is not hindered from doing so. Entropy is the quantitative measure of that kind of spontaneous process: how much energy has flowed from being localized to becoming more widely spread out (at a specific temperature).

The amount of energy spreading when an amount of heat \(\Delta Q\) is introduced into a system at temperature \(T\) is proportional to \(\Delta Q/T\).

Consequence of the principle: Entropy is a nondecreasing function of time.
  1. Gases expand but never contract: in a large volume, there are many more accessible micro states.

  2. Mixing is irreversible.

  3. Heat flows from hot to cold.

  4. Explanation of Le Chatelier’s Principle: for an endothermic reaction, equilibrium shifts to the right as temperature increases. Example: \(\ce{ H2 → 2H }\).

14.1. References

[UE] Understanding entropy. This is a very well written article that gives a lot of insight into the nature of entropy based on the number of accessible micro states.

[TE] Teaching entropy (Lambert)

15. Exoplanets

15.1. The calculation

  1. Parallax method for determining distance to star — use earth’s orbit as baseline

  2. Use absolute luminosity from previous step plus observed luminosity to calculate mass of star.

  3. Use Kepler-Newton relation to determine the orbital radius.

  4. Use Doppler shift measured and the radial velocity method to determine the force the start exerts on the planet. Then use Universal Law of Gravitation to determine mass of planet.

  5. Use the Stefan-Boltzmann law to determine the radius of the star.

  6. Use the luminosity drop when the planet passes over the stars disk and the relation \(\Delta L = r^2/R^2\) to determine the radius of the planet.

mass luminosity relation

16. Air Pressure

Imagine a spherical chamber 20 cm in diameter composed of two hemispheres sealed by a gasket. The surface area of the chamber is \(4\pi 10^2\) cm2, or about 1200 cm2. In SI units, the surface area is \(2.4\times 10^{-1}\) m2. One atmosphere of pressure is 101325 Pascals (Newton/m2), that is \(1.01\times 10^5\) N/m2. Consider the force directed towards the right on the left hemisphere. This is the same as the pressure times the area of the equatorial disk. The area of the equatorial disk is \(\pi\times10^2\) cm2, or about \(3\times 10^{-2}\) m2. Thus the force is about \(3\times 10^3\) Newtons.