Question

What is Callendar's Equation and how can we derived it from volume of expansivity?

Answer 1

The equation of state for an ideal gas is

pV = RT

where p is gas pressure, V is volume, 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 C_V (j/(^oK mole)), then we may set dq = C_V dT. It follows, in this case, that

du = CV 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 = R 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 C_p, then dq = C_p dT, and, from 2 (with 3),

C_p dT = C_V dT + R dT

Simplifying gives an important constitutive relationship between C_V, C_p, 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 = C_V 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 RT, and the second term by pV). Further, from1,

p dV + V dp = R dT

or, equivalently,

dp/p + dV/V = dT/T

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

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 = (C_V/R)(dp/p + dV/V) + dV/V, or

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

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

-dp/p = dV/V

which, upon integration, yields

p0/p = (V/V0)

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

Case 2:

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

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

Proceeding as before produces the result that

	/( - 1)
p/p0 = (T/T0)	8.

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

dS = dq/T

9.

where dS is the differential entropy change.

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

S = CV ln(T/T0)

10.

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

S = Cp ln(T/T0)

11.

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

S = R ln(V/V0)

12.

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

S = const.

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

Solution:
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 + gh = p0

13.

where p is the pressure of air at height h above the earth's surface, 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 + g dh = 0

13a.

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 = /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 , we have = 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)h

14.

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 h = 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

or

T = 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.