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Fermi's Piano Tuner Problem

How Old is Old?

If the Terrestrial Poles were to Melt...

Sunlight Exerts Pressure

Falling Eastward

What if an Asteroid Hit the Earth

Using a Jeep to Estimate the Energy in Gasoline

How do Police Radars really work?

How "Fast" is the Speed of Light?

How Long is a Light Year?

How Big is a Trillion?

"Seeing" the Earth, Moon, and Sun to Scale

Of Stars and Drops of Water

If I Were to Build a Model of the Cosmos...

A Number Trick

Designing a High Altitude Balloon

Pressure in the Vicinity of a Lunar Astronaut Space Suit due to Outgassing of Coolant Water

Calendar Calculations

Telling Time by the Stars - Sidereal Time

Fields, an Heuristic Approach

The Irrationality of

The Irrationality of

The Number (i)i

Estimating the Temperature of a Flat Plate in Low Earth Orbit

Proving that (p)1/n is Irrational when p is a Prime and n>1

The Transcendentality of

Ideal Gases under Constant Volume, Constant Pressure, Constant Temperature and Adiabatic Conditions

Maxwell's Equations: The Vector and Scalar Potentials

A Possible Scalar Term Describing Energy Density in the Gravitational Field

A Proposed Relativistic, Thermodynamic Four-Vector

Motivational Argument for the Expression-eix=cosx+isinx

Another Motivational Argument for the Expression-eix=cosx+isinx
Calculating the Energy from Sunlight over a 12 hour period
Calculating the Energy from Sunlight over actual full day
Perfect Numbers-A Case Study
Gravitation Inside a Uniform Hollow Sphere
Further note on Gravitation Inside a Uniform Hollow Sphere
Pythagorean Triples
Black Holes and Point Set Topology
Additional Notes on Black Holes and Point Set Topology
Field Equations and Equations of Motion (General Relativity)
The observer in modern physics
A Note on the Centrifugal and Coriolis Accelerations as Pseudo Accelerations - PDF File
On Expansion of the Universe - PDF File

Ideal Gases under Constant Volume, Constant Pressure, Constant Temperature, & Adiabatic Conditions

Note to the student: The following section is a reduction of college notes I made in introductory thermodynamics. It does not read as easily as the preceding sections. I include it here because, for me, it represented a significant unification of the ideas presented in the text and during lecture. The first year college student will certainly find it useful.

The equation of state for an ideal gas is

pV = muRT


where p is gas pressure, V is volume, mu is the number of moles, R is the universal gas constant (= 8.3144 j/(oK mole)), and T is the absolute temperature. The first law of thermodynamics, the conservation of energy, may be written in differential form as

dq = du + p dV


where dq is a thermal energy input to the gas, du is a change in the internal energy of the gas, and p dV is the work done by the gas in expanding through the change in volume dV.


Constant Volume Process

If V = const., then dV = 0, and, from 2, dq = du; i.e., all the thermal input to the gas goes into internal energy of the gas. We should expect a temperature rise. If the gas has a specific heat at constant volume of CV (j/(oK mole)), then we may set dq = muCV dT. It follows, in this case, that

du = muCV dT


Since du was initially unspecified, we are free to choose its mathematical form. Equation 2 will be retained for du throughout the remainder of the cases.


Constant Pressure Process

If p = const., then dp = 0, and, from 1, p dV = muR dT; i.e., the work done by the gas in expanding through the differential volume dV is directly proportional to the temperature change dT. If the gas has a specific heat at constant pressure of Cp, then dq = muCp dT, and, from 2 (with 3),

muCp dT = muCV dT + muR dT

Simplifying gives an important constitutive relationship between CV, Cp, and R, namely:

Cp = CV + R



Constant Temperature Process

If T = const., then dT = 0, and, from 1, d(pV) = 0, i.e., pressure and volume are inversely proportional. Further, from 2, dq = p dV; i.e., there is no change in internal energy (from 3, du = 0), and all the thermal input to the gas goes into the work of expansion.


Adiabatic Process

If q = const, then dq = 0, and, from 2 (with 3), 0 = muCV dT + p dV; i.e., internal energy of the gas might be reduced in favor of expansion, or vice versa. This expression may be written in an equivalent form as

0 = (CV/R)(dT/T) + dV/V


(division of the first term by muRT, and the second term by pV). Further, from 1,

p dV + V dp = muR dT

or, equivalently,

dp/p + dV/V = dT/T


(division of the Left Hand Side by pV, and the Right Hand Side by muRT).

Equations 5 and 6 may be used to develop relationships between p and V, or p and T:


Case 1:

To eliminate T, use 6 in 5 for dT/T to obtain 0 = (CV/R)(dp/p + dV/V) + dV/V, or

-(CV/R) dp/p = (1 + CV/R) dV/V


Using 4, we may write CV/R = CV/(Cp - CV) = 1/(y- 1) where y = Cp/CV, the ratio of the specific heats (>1; in fact, approximately 1.4 for air at STP). Thus, equation 6a becomes (after simplification)

-dp/p = ydV/V

which, upon integration, yields

p0/p = (V/V0)y


i.e., the pressure varies inversely as the volume raised to the power y.


Case 2:

To eliminate V, use equation 6 to write dV/V = dT/T - dp/p, and substitute for dV/V in eq. 5

(CV/R + 1) dT/T = dp/p

Proceeding as before produces the result that

y/(y - 1)

p/p0 = (T/T0)


Entropy changes may be calculated for each of the above thermodynamic processes. The definition of entropy is

dS = dq/T


where dS is the differential entropy change.

For constant volume processes, dq = muCV dT, so that dS = muCV dT/T, and

deltaS = muCV ln(T/T0)


For constant pressure processes, dq = muCp dT, so that dS = muCp dT/T, and

deltaS = muCp ln(T/T0)


For constant temperature processes, dq = p dV, so that dS = p dV/T = muR dV/V, and

deltaS = muR ln(V/V0)


For adiabatic processes, dq = 0, so that ds = 0, and

S = const.


Estimate the dry adiabatic lapse rate for an ascending parcel of air near the earth's surface.

The dry adiabatic lapse rate for an ascending parcel of air near the earth's surface may be estimated from the above expressions. (For comparison, the published value is 5.5 oF per thousand feet). Let us begin with Bernoulli's equation

p + rhogh = p0


where p is the pressure of air at height h above the earth's surface, rho is the density of air in kg/m3, g is the gravitational acceleration, and p0 is atmospheric pressure at the earth's surface. Differentiating this expression once, we get

dp + rhog dh = 0


We now write the equation of state for an ideal gas as p = nkT, where n is the number density of the gas in m-3, and k is Boltzmann's constant. We may further write n = rho/M where M is the average molecular mass of air in kg. Using this expression in the ideal gas equation of state and solving for rho, we have rho = pM/kT. Substituting this result into 13a gives

dp + (pMg/kT) dh = 0

We may now separate variables and integrate. In so doing, it is customary to assume that the variation of T with h may be ignored (isothermal approximation). Thus

dp/p = -(Mg/kT) dh

or p/p0 = e-(Mg/kT)deltah


The term kT/Mg is often called the scale height of the atmosphere (i.e., that height at which the calculated pressure drops to 1/e of its initial value; in an isothermal atmosphere, this number has physical meaning). At the earth's surface, at STP (Standard Temperature and Presure, 1 atm.), we have

M = 4.81 x 10-26 kg
g = 9.81 m/sec2
k = 1.38 x 10-23 j/oK
T = 273 oK

so that the scale height is 7.99 km = 2.53 x 104 ft. If we now set p0 = 1 atm, then at deltah = 1,000 ft., we find

p = 9.61 x 10-1 atm.

We may use this pressure in equation 8 as an estimate of atmospheric pressure (even though eq. 8 pertains to an adiabatic atmosphere rather than an isothermal one), and calculate the temperature at 1,000 ft. When we do so, we find

T = 270 oK


deltaT = 3 oK = 5.4 oF


This value is satisfactorily close to the book value of 5.5 oF per 1,000 ft. to meet our needs.

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