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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, 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 =
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 = ![]() ![]() ![]() Simplifying gives an important constitutive relationship between CV, Cp, 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 = (division
of the first term by (division
of the Left Hand Side by pV, and the Right Hand Side by 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 Using
4, we may write CV/R = CV/(Cp
- CV) = 1/( -dp/p
=
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 (CV/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 =
For constant
pressure processes, dq =
For constant
temperature processes, dq = p dV, so that dS = p dV/T =
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, 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 = 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/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 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
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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