Outline
• Ideal gas equation
• Compressibility factor
PvT relations for gases
• Tables (of state) vs. equations of state for property data
• Gas vs. vapor i.e. gas above Tcr, no condensation
• Experiments by Boyle (1662), Charles and Gay-Lussac
(1802) found at low pressures, volume of gas prop. to T
T 
P = R   → Pυ = RT
υ 
Ideal gas equation of state
obeyed by ideal gases
• R proportionality constant – gas constant (different for
each gas); Ru – universal gas constant
8.314kJ / kmol − K

R = Ru / M where Ru = 1.986 Btu / lbmole − R
⋮

Boyle’s Law
1
Charles and Gay-Lussac’s Law
Ideal gas equation of state
• Molar mass, M, mass of 1 mole of substance in
grams; gmol=gram mole; or mass of 1 kmole in
kg; or mass of 1 lbmole in lbm
• 1 mole of substance contains Avogadro’s no. of
molecules (6.023X1023 molecules/mole)
• Ex. MN2 =28 so 1 kmol of N2 is 28 kg or 1 lbmol
of N2 is 28 lbm; M=m/N, N is mole no.
PV = mRT → V = mυ
PV = NRuT → mR = MNR = NRu
Pυ = RuT → V = Nυ (bar per kmole)
PV
PV
1 1
= 2 2
T1
T2
What is an ideal gas (IG)?
• IG is imaginary substance that obeys ideal
gas equation of state e.g. Pv=RT
• It assumes zero molecular volume and no
attractive forces between molecules
• It works well at low densities i.e. low P,
high T
• How low is low and how high is high?
• Introduce compressibility factor
2
Is water vapor an IG?
• Below 10 kPa at all T
yes
• At high pressure, near
CP and SV line no
• In AC applications,
vapor pressure is low
so water vapor ~ IG
• In steam powerplants,
high pressure so IG is
bad
Compressibility factor
• Measures deviations from IG behavior
Z=
Pυ 1 for IG
=
;
RT >< for real gas
Pυ = ZRT ⇒ Z =
υ actual
RT
w/υideal =
υideal
P
• Low P and high T? Relative to what?
• Experimental observation: Different gases behave
differently at given P, T but behave similarly at P, T
normalized by Pcr, Tcr
• Z is approx. same for all gases at same TR=T/Tcr,
PR=P/Pcr, principle of corresponding states
Generalized compressibility chart
PR<<1 IG for all TR
TR>>2 IG for all PR, except PR>>1
Worst near CP e.g. TR=PR=1
3
Example
• Determine the specific volume of refrigerant-134a at
1MPa and 50C using (a) ideal gas equation of state and
(b) generalized compressibility chart. Compare values
obtained to actual value of 0.02171m3/kg and determine
the error involved in each case.
A − 1: R = 0.8015kPa − m3 / kg − K ; Pcr = 4.067 MPa; Tcr = 374.3K
(a ) υ = RT / P = 0.02632m3 / kg (21.2%)
(b) PR = P / Pcr = 1MPa / 4.067 MPa = 0.246;
TR = T / Tcr = 323K / 374.3K = 0.863
∴ Z = 0.84;υ = Zυideal = (0.84)(0.02632) = 0.02211m3 / kg (2%)
Summary
• Under what conditions is the ideal gas
assumption suitable for real gases?
• What is the difference between R and Ru? How
are these two related?
• What is the difference between mass and molar
mass?
• What is the physical significance of the
compressibility factor, Z?
• How are reduced pressure and reduced
temperature defined?
Outline
• Examples
• Properties of ideal gases
4
Example
• A 0.5m3 rigid tank containing hydrogen at 20C and
600kPa is connected by a valve to another 0.5m3 rigid
tank that holds hydrogen at 30C and 150kPa. Now the
valve is opened and the system is allowed to reach
thermal equilibrium with the surroundings, which are at
15C. Determine the final pressure in the tank.
T
H2
V=0.5m3
T=20C
P=600kPa
X
H2
V=0.5m3
T=30C
P=150kPa
15C
Surroundings
Solution
V = V A + VB = 0.5 + 0.5 = 1m3
A-1 for gas constant
 PV 
(600kPa )(0.5m 3 )
mA =  1  =
= 0.248kg
RT
(4.124
kPa
− m 3 / kg − K )(293K )
 1 A
 PV 
(150kPa )(0.5m 3 )
mB =  1  =
= 0.060kg
RT
(4.124
kPa
− m 3 / kg − K )(303K )
 1 B
m = m A + mB = 0.248 + 0.060 = 0.308kg
P = mRT2 / V
= (0.308kg )(4.124kPa − m3 / kg − K )(288 K ) /1.0m3 = 365.8kPa
Example
• A 20m3 tank contains nitrogen at 25C and
800kPa. Some nitrogen is allowed to escape
until the pressure in the tank drops to 600kPa. If
the temperature at this point is 20C, determine
the amount of nitrogen that has escaped.
N2
800kPa
25C
20m3
5
Solution
V = VA + VB = 0.5 + 0.5 = 1m3
m1 =
PV
(800kPa )(20m 3 )
1
=
= 180.9kg
RT1 (0.2968kPa − m 3 / kg − K )(298 K )
m2 =
PV
(600kPa )(20m 3 )
2
=
= 138.0kg
RT2 (0.2968kPa − m 3 / kg − K )(293K )
∆m = m1 − m2 = 180.9 − 138.0 = 42.9kg
Internal energy, enthalpy, and
specific heats – ideal gases
Pυ = RT
Can show for IG: u = u (T )
Recall h = u + Pυ ∴ h = h(T )
Hence, CP , Cv = f (T ) ∴
2
du = Cv (T ) dT → ∆u = ∫ Cv (T )dT
1
2
dh = CP (T )dT → ∆h = ∫ CP (T )dT
1
How do specific heats vary with temperature?
Ideal gas constant-pressure
specific heats for some gases
• Specific heats of
complex molecules (2
or more atoms)
increase with T
• Smooth variation with
T and is roughly
constant over small T
range
• For monotomic
gases, specific heat is
constant for all T
6
Evaluating the integrals – 3 choices
• Choice 1: Polynomial curve fits from measured data (A-21)
cP
= α + β T + γ T 2 + δ T 3 + ε T 4 ; T ( K ); cP (kJ / kmol − K )
R
• Choice 2: Ideal gas tables (A-22/23) with u=h=0 @ 0K as
reference state
2
T2
T1
1
0
0
∆u = ∫ Cv (T )dT = ∫ Cv (T )dT − ∫ Cv (T ) dT = u (T2 ) − u (T1 )
• Choice 3: Assume constant values (A-20)
2
∆u = u2 − u1 = ∫ Cv (T ) dT = Cv ,ave (T2 − T1 )
1
Specific heat relations for ideal
gases
h = u + Pυ = u + RT ∴ dh = du + RdT
→ CP = Cv + R also CP / Cv ≡ k = 1 + R / Cv
k is called specific heat ratio
k → 1.667 for monatomic gases
k → 1.4 for diatomic gases (air)
above for room temperature
Example
• A 3m3 rigid tank contains hydrogen at 250kPa
and 500K. The gas is now cooled until its
temperature drops to 300K. Determine the (a)
final pressure in the tank and (b) amount of heat
transfer.
• Assumptions: (1) Hydrogen modeled as ideal
gas since it is at temperature and low pressure
relative to its critical point values of -240C and
1.30MPa; (2) Tank is stationary.
7
Solution
(a )
PV
PV
T
300 K
1
(250kPa ) = 150kPa
= 2 → P2 = 2 P1 =
T1
T2
T1
500 K
(b)Q − W = ∆E = ∆U = m ( u2 − u1 )
W = 0 (rigid tank, no shafts or wires)
u2 − u1 ≅ Cv (T2 − T1 ) (use Cv ,ave @ 400 K = 10.352kJ / kg − K ; A − 2)
m=
PV
(250kPa )(3.0m 3 )
1
=
= 0.3637kg (A -1)
RT1 (4.124kPa − m 3 / kg − K )(500 K )
∴ Q = (0.3637kg )(10.352kJ / kg − K )(500 − 300) K = 753.0kJ
8