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Ideal Gases under Constant Volume, Constant Pressure, Constant Temperature, & Adiabatic ConditionsNote 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 where p is gas pressure, V is volume, is the number of moles, R is the universal gas constant (= 8.3144 j/(^{o}K mole)), and T is the absolute temperature. The first law of thermodynamics, the conservation of energy, may be written in differential form as 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/(^{o}K mole)), then we may set dq = C_{V} dT. It follows, in this case, that 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:
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 (division of the first term by RT, and the second term by pV). Further, from 1, (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 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
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 Entropy changes may be calculated for each of the above thermodynamic processes. The definition of entropy is
where dS is the differential entropy change. For constant volume processes, dq = C_{V} dT, so that dS = C_{V} dT/T, and
For constant pressure processes, dq = C_{p} dT, so that dS = C_{p} dT/T, and
For constant temperature processes, dq = p dV, so that dS = p dV/T = R dV/V, and
For adiabatic processes, dq = 0, so that ds = 0, and S =
const.
Problem: Solution:
where p is the pressure of air at height h above the earth's surface, is the density of air in kg/m^{3}, g is the gravitational acceleration, and p_{0} is atmospheric pressure at the earth's surface. Differentiating this expression once, we get 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
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/sec^{2} k = 1.38 x 10^{-23} j/^{o}K T = 273 ^{o}K so that the scale height is 7.99 km = 2.53 x 10^{4} ft. If we now set p_{0} = 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 ^{o}K
This value is satisfactorily close to the book value of 5.5 ^{o}F per 1,000 ft. to meet our needs. |
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