Summary of Topics from Introductory Problems on Thermal Energy and Fluids


Symbols used:

Note that symbols might not be correctly represented by your browser.  For this reason the Greek letters will be spelled out, with ` in front of the spelling.  You should substitute the appropriate symbol when making notes.

Suggestion:  If you want to memorize formulas, the formula at the beginning of each bulleted paragraph is probably worth the effort.   Those formulas that aren't worth the effort are enclosed in parentheses.

Fluids:

Thermal Energy Transfers and Materials:

Kinetic Theory of Gases:

One particle mass m, vel. v in cylinder length L, moving parallel to axis of cylinder exerts average force F = `d(mv) / `dt = 2 m v / ( 2 L / v) = m v^2 / L on either end.

pressure = F / A = m v^2 / (L * A) = m v^2 / V, where V is volume.

Noting

(n is the number of moles. To find KE per particle divide by the number of moles to get the KE of 1 mole, then by Avagadro's Number to get the KE of 1 particle.)

(note for General College and especially University Physics students:   the KE referred to here is the KE of a single particle at at the rms velocity)

Gas Laws:

V2 / V1 = T2 / T1:  If P and n are constant then V is proportional to T and you can use 'straight ratios' V2 / V1 = T2 / T1 to calculate desired temp or volume.

V2 / V1 = P2 / P1:  If V and n are constant then P is proportional to T and you can use 'straight ratios' V2 / V1 = P2 / P1 to calculate desired pressure or volume.

P2 / P1 = V1 / V2:  If T and n are constant then P is inversely proportional to V and you can use inverse ratios P2 / P1 = V1 / V2 (note that one of the ratios is 'upside down') to calculate desired temp or volume.

If you know three of the four quantities P, V, n and T you can find the fourth.

Thermodynamics:

efficiency = work done by system / thermal energy transfered into system, or

efficiency = `dW / `dQin .

The maximum possible efficiency of any such system is (Th – Tc) / Th, where Th and Tc are the temperatures of the hot and cold sources.  This efficiency is theoretically achievable only by a Carnot Cycle, which consists of two adiabatic processes bounded by two isothermals.

Summary of Topics from Introductory Problems on Waves and Optics


Waves and Optics:

T = 1 / f:  The period T is the time required for a peak-to-peak cycle, and is equal to the reciprocal of the frequency: T = 1 / f.

`omega = 2 `pi f:   The angular frequency `omega is the velocity of the point moving around the reference circle which models the simple harmonic motion of a single point as the wave passes. Since there are 2 `pi radians in a circle, `omega = 2 `pi f.

E = k Q / r^2:  If the charge distribution is spherically symmetric, then the electric field due to that distribution will be spherically symmetric about the same central point. The field will be directed radially outward from the center and will hence be perpendicular to the surface of any concentric sphere. Thus for any such sphere we have uniform electric field E = 4 `pi k Q / ( 4 `pi r^2) = k Q / r^2, where Q is the total charge enclosed by the sphere. Note that this agrees with Coulomb's Law in the case of a point charge q.

E = 2 k `lambda / r:  If a charge distribution is cylindrically symmetric, then the electric field due to that distribution will be cylindrically symmetric about the central axis of the distribution. The field will be directed radially outward from the central axis and will hence be perpendicular to the curved surface of any coaxial cylinder. Thus if `lambda is the amount of charge per unit length enclosed by the cylinder, a cylinder of radius r and length L will enclose charge Q = `lambda * L. Since the curved surface of the cylinder will have area A = 2 `pi r * L, the field at the surface of the cylinder will be E = 4 `pi k Q / r^2 = 4 `pi k ( `lambda * L) / (2 `pi r * L) = 2 k `lambda / r.

E = 2 `pi k `sigma:  If charge is uniformly distributed over a large plane area, then symmetry arguments show that near that plane but not close to its edges the electric field due to the distribution is very nearly perpendicular to the plane. It follows that if the charge per unit area is `sigma, then a rectangular box with its central axis oriented perpendicular to the plane, with the box intersecting the plane and having cross-sectional area A perpendicular to the plane, will contain charge Q = `sigma * A. The electrostatic flux will exit this box through its two ends and not through its sides, so the flux will be E * ( 2 A). We therefore have E ( 2 A) = 4 `pi k ( `sigma * A) and E = 4 `pi k ( `sigma * A ) / ( 2 A) = 2 `pi k `sigma. Note that as long as we remain near the plane, in the sense already specified, the electric field remains constant.

Work, Energy, Potential Difference, Power:

Conduction by Charge Carriers:

Ohm's Law and Circuits:

Capacitors and Capacitance:

Magnetism:

Summary of Topics from Introductory Problems on Modern Physics


Quantization of Energy in Electromagnetic Radiation:

Wave Properties of Particles:

Quantization of Orbital Energies in Atoms:

Conversion between Mass and Energy and Nuclear Decay Modes:

Special Relativity:

Brief Notes about Particle Physics: