1. FAYETTEVILLE STATE UNIVERSITY COLLEGE OF BASIC AND APPLIED SCIENCES DEPARTMENT OF NATURAL SCIENCES CHEM Fall 2015
Chapter 1,7 Gases

2. Characteristics of Gases
Unlike liquids and solids, they
Expand to fill their containers.
Are highly compressible.
Have extremely low densities.

3. Pressure F P = A Pressure is the amount of force applied to an area.
Atmospheric pressure is the weight of air per unit of area.

4. Units of Pressure Pascals Bar 1 Pa = 1 N/m2 1 bar = 105 Pa = 100 kPa
1 kPa = Pa

5. Units of Pressure mm Hg or torr
These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury.
760 mm Hg
Atmosphere
1.00 atm = 760 torr

6. Manometer
Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

7. Standard Pressure Normal atmospheric pressure at sea level.
It is equal to
1.00 atm
760 torr (760 mm Hg)
1 atm = kPa
1 bar =105 Pa

8. Boyle’s Law
The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

9. Boyle’s Law

10. As p and V are inversely proportional
A plot of V versus P results in a curve.
Since
V = k (1/p)
This means that a plot of V versus 1/p will be a straight line.
pV = k

11. Charles’s Law
The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.
i.e., V T
= k
A plot of V versus T will be a straight line.

12. Avogadro’s Law
The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.
Mathematically, this means
V = kn

13. Ideal-Gas Equation V  nT P So far we’ve seen that
V  1/P (Boyle’s law)
V  T (Charles’s law)
V  n (Avogadro’s law)
Combining these, we get
V  nT P

14. PV = nRT Ideal-Gas Equation nT P V  nT P V = R The relationship
then becomes
PV = nRT

15. Ideal-Gas Equation
The constant of proportionality is known as R, the gas constant.
STP Conditions:
Standard Temperature and Pressure
Ts = 00 C = 273 K
Ps = 1 atm
At STP:
1 mol o f gas occupy a volume of
V = 22.4 L
R = Ps * Vs/(1mol*Ts)
R = 1 atm * 22.4 L/(1mol * 273K) = L-atm/mol-K

16. Densities of Gases n P V = RT
If we divide both sides of the ideal-gas equation by V and by RT, we get n V P RT =

17. Density and Molar Mass of Gases
We know that
moles  molecular mass = mass
n   = m
So multiplying both sides by the molecular mass ( ) gives
mRT PV
M = P RT m V = P RT m V =
d =
Mass  volume = density

18. EXAMPLE #1 M = = 44.5 g/mol 0.896 g x0.082 x 301 K
A gas sample weigh 0.896g and occupy a volume of 524 ml
at pressure of 720 mm Hg and temperature of 28 0C.
V = L
T = = 301 K
P = 720mm Hg x 1atm / 760 mm Hg = atm
Calculate the molar mass?
0.896 g x0.082 x 301 K
M =
atm x L
= 44.5 g/mol

19. Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases equals the sum of the pressures that each gas would exert if it were present alone.
In other words,
Ptotal = P1 + P2 + P3 + …

20. Partial Pressures
When one collects a gas over water, there is water vapor mixed in with the gas.
To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

21. Ideal Gas
This is a model that aids in our understanding of what happens to gas particles as environmental conditions change.

22. Correction to the Ideal Gas Behavior
Compression Factor (Z = pVm/RT)
Virial Expansion
Van der Waal Equation

23. Real Gases and compression Factor (z = pVm/RT = Vm/Vmideal)
In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure (Z=1).

24. The Boyle’s Temperature
z = pVm/RT = f(T,p)

25. Virial Expansion pVm = RT(1 + B'p + C'p2 +.
s =1 + q + q2 + q3+……
pVm = RT(1 + B'p + C'p2 +.
B' and C' The first and second
virial coefficients
Also expressed as: pVm = RT(1 + B/Vm + C/Vm2 +….
The slope of Z, (dZ/dp)=0, perfect gas, Z = 1
The Boyle’s Temperature TB (real gas behaves ideal)
If B' and C' diff from “0” Real Gas

26. Deviations from Ideal Behavior
The assumptions made in the kinetic-molecular model break down at high pressure and/or low temperature.

27. Corrections for Nonideal Behavior
The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account.
The corrected ideal-gas equation is known as the van der Waals equation.

28. Correction to the Ideal Equation
Correction to the volume nb, b correction
b dm3/mol (L/mol)
Correction to the Pression a/Vm2
a (dm6-atm)/mol2 or (L2-atm)/mol2
V – nb Volume available to the gas

29. The van der Waals Equation
) (V - nb) = nRT
n2a V2
(p +

30. Solving the Van der Waals Equation Example#2
Example#1:calculate the pressure exerted by 15g of H2 in a volume
of 5dm3 at 300K using the van der Waals equation.
Use a = (dm6-atm)/mol2; b = dm3/mol
Vm = V/n = 5*2.016/15 = dm3/mol
p = RT/(Vm - b) - a/Vm2
p = [0.082*300]/( ) /(0.672)2 = 37.6 atm

31. Solving the Van der Waals Equation for Vm iteratively Example#3
Example#2: Calculate Vm for H2 at p = 40.0 atm, T = 300K, use the a and b values given above.
Vm = RT/(p + a/Vm2) + b
Calculate the Vm guess value using ideal gas equation:
Vm(1) = RT/p = 0.082*300/40.0 = 0.615
Vm(2) = RT/(p + a/Vm(x)2) + b =
Vm(2) = 0.082*300/( /(0.615) = dm3/mol
Vm(3) = 0.082*300/( /(0.6278) = dm3/mol
Vm(4) = 0.082*300/( /(0.6282) = dm3/mol
Vm = Vm(4) = dm3/mol

32. Isothermal Compression of a Gas
T<Tc
ℓ→v
Equil.
Critical Point
T=Tc
Vg=Vℓ
Supercritical fluid
T>Tc
Compressed
Gas Isotherms
T<Tc
Gas

33. Van der Waals Isotherm vl vg Liquid Compressed Gas Isotherms
Van der Waals Compression Described by the van der Gas Equation
Liquid
Super
critical
Fluid
Van der Waals Isotherm
Liquid-Gas Equil
Gas
Unrealistic behavior

34. Compression of a Real Gas Real Gas Isotherm
Supercritical Fluid
Fig Experimental
isotherms of carbon dioxide at
several temperatures. The
‘critical isotherm’, the isotherm
at the critical temperature, is at 31.04°C. The critical point is
marked with a star
Compressed Gas
Isotherms
Liquid
Liquid –vapor
Equilibrium vl vg

35. Compression of
van der Waals Gas

36. Useful Math http://www.sosmath.com/
Chain Rule Derivatives

37. Solving Van der Waals Equation at the Critical Point (Vc, pc van der Waals, Tc)
At the critical point(inflexion point) of the isotherms:
(p/V)T = 0 and (2p/V2)T = 0
p = RT/(V - b) - a/V2, V is the molar volume of 1 mol ( Vm = V)
(p/V)c = -RTc/(Vc - b)2 + 2a/Vc3 = 0  RTc/(Vc - b)2 = 2a/Vc3
and
(2p/V2)c = 2RTc/(Vc - b)3 - 6a/Vc4 = 0  2RTc/(Vc - b)3 = 6a/Vc4
Take the ratio of the two expressions rearrange and solve:
Vc = 3b b= Vc /3
pc = a/27b2 a= 3Vc2Pc
Tc = 8a/27Rb

38. 8Tr Vr -1 pr = 3 Vr2 - Correspondent States Vc = 3b Tc = 8a/27Rb
pc = a/27b2
Reduced Variables
pr = p/pc, Vr = V/Vc, Tr = T/Tc
Reduced Equation
8Tr
Vr -1
pr = 3
Vr2 -

39. The principle of Corresponding States
Vr=VA/VAc Vr=VB/VBc
Tr=TA/TAc Tr=TB/TBc
pr=pA/pAc pr=pB/pBc
If two gases are at the same Vr and Tr they will be at the same reduced pressure or
If VA/Vac = VB/VBc and TA/Tac = TB/TBc then pA/pAc = pB/pBc
Therefore, the two gases are at same corresponding state
Gas A
Vr, Tr
Gas B
Vr, Tr
Gas A and B will
be at the same Pr

40. The principle of Corresponding States
pr = 8Tr/(Vr -1) – 3/Vr2
At the critical point:
pr = p/pc = Vr = V/Vc = Tr = T/Tc = 1
Fig Van der Waals isotherms at several values of T/Tc. Compare these curves with the Van der Waals isotherms. The van der Waals loops are normally replaced by horizontal straight lines. The critical isotherm is the isotherm for T/Tc = 1.
EXPLORATION Calculate the molar volume of chlorine gas on the basis of the van der Waals equation of state at 250 K and 150 kPa and calculate the percentage difference from the value predicted by the perfect gas equation.

41. Example#4: Find the p and T at which the Ar gas is a the same corresponding state as N2.
For N For Ar
T = K Tc = K Tc = T = ?
p = 1 atm pc = K pc = p = ?
Tr = T/Tc = /44.44 = 6.709
pr = p/pc = 1/33.54 =
Two Gases are a the same corresponding state if the have the
same Vr, Tr and pr. Therefore N2 and Ar will be in the same
corresponding state if they have the same Tr and pr values
Ar will be at the same corresponding state as N2 when:
p = prpc = *48.00 = atm
T = Tr*Tc = 6.709* = 1,011 K