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
= RT
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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
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 = CV
dT. It follows, in this case, that
du
= CV
dT
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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 Cp, then
dq = Cp
dT, and, from 2 (with 3),
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 = CV
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,
p
dV + V dp = R
dT
or,
equivalently,
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dp/p
+ dV/V = dT/T |
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(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 = (CV/R)(dp/p + dV/V) + dV/V, or
-(CV/R)
dp/p = (1 + CV/R) dV/V
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Using
4, we may write CV/R = CV/(Cp
- CV) = 1/(-
1) where
= 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
= dV/V
which,
upon integration, yields
p0/p
= (V/V0)
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7.
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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
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/(
- 1) |
p/p0
= (T/T0) |
8. |
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 = CV
dT, so that dS = CV
dT/T, and
S
= CV
ln(T/T0)
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10.
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For
constant pressure processes, dq = Cp
dT, so that dS = Cp
dT/T, and
S
= Cp
ln(T/T0)
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11.
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For
constant temperature processes, dq = p dV, so that dS = p dV/T = R
dV/V, and
S
= R
ln(V/V0)
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12.
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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
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13.
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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
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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
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14.
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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
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T
= 3 oK = 5.4 oF
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15.
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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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