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Chapter 5 Gases
5.1 Substances that exist s gases
5.2 Pressure of the gas
5.3 The gas laws
5.4 Ideal gas equation
5.5 Gas stoichiometry
5.6 Dalton’s Law of Partial Pressures
p215: 5.1, 5.2, 5.3, 5.8, 5.14, 5.22
p216: 5.24, 5.32, 5.36, 5.38, 5.40, 5.44, 5.48, 5.50, 5.94
p217: 5.52, 5.54, 5.64, 5.66, 5.68, 5.70
5.1 substances that exist s gases
Elements that exist as gases at 250C and 1 atmosphere
Physical Characteristics of Gases
Gases assume the volume and shape of their containers.
Gases are the most compressible state of matter.
Gases will mix evenly and completely when confined to the same container.
Gases have much lower densities than liquids and solids.
5.2 Pressure of Gases and its Units
Pressure is defined as the force applied per unit are
The SI unit of pressure is Pascal (Pa) define as one Newton per square meter ( 1Pa = N/m2)
Standard atmospheric pressure, the pressure that supports a column of mercury exactly 760 mm high at 0 oC at sea level.
Measured using a Barometer! - A device that can weigh the atmosphere above us!
a Barometer! - A device that can weigh the atmosphere above us!
A simple manometer, a device for measuring the pressure of a gas in a container
Common Units of Pressure
Unit Atmospheric Pressure Scientific Field Used
Pascal (Pa); 1.01325 x 105 Pa SI unit; physics,
kilopascal (kPa) 101.325 kPa chemistry
Atmosphere (atm) 1 atm Chemistry
Millimeters of mercury 760 mmHg Chemistry, medicine
Torr 760 torr Chemistry
Pounds per square inch 14.7 lb/in2 Engineering
(psl or lb/in2)
Bar 1.01325 bar Meteorology, chemistry
Remember the conversions for pressure:
760 mm Hg = 760 torr
1 atm = 760 mm Hg
760 mm Hg = 101.325 Pa
Convert 2.3 atm to torr:
2.0 atm x760 torr = 1520 torr
Worked Example 5.1
Boyle’s Law , V - P relationship
Charles Law , V - T- relationship
Avogadro’s Law ,V and Amount
Boyle’s Law : P - V relationship
Pressure is inversely proportional to Volume
P = or V = or
Change of Conditions Problems
if n and T are constant
P1V1 = k P2V2 = k’
k = k’
P1V1 = P2V2
When you double the pressure on a gas,
the volume is cut in half (as long as the
temperature and amount of gas do not change)
Example A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?
V1, P1, P2
V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm
P1 ∙ V1 = P2 ∙ V2
since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does
A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2780 mL, what was it originally?
V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm
P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)
since P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does
volume is directly proportional to temperature
constant P and amount of gas
as T increases, V also increases
Kelvin T = Celsius T + 273
V = constant x T
if T measured in Kelvin
A gas has a volume of 2.57 L at 0.00°C. What was the temperature at 2.80 L?
T(K) = t(°C) + 273.15,
V1, V2, T2
V1 =2.57 L, V2 = 2.80 L, t2 = 0.00°C
t1, K and °C
since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does
volume directly proportional to the number of gas molecules
V = constant x n
constant P and T
more gas molecules = larger volume
count number of gas molecules by moles
equal volumes of gases contain equal numbers of molecules
the gas doesn’t matter
mol added = n2 – n1,
V1, V2, n1
A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?
V1 =4.65 L, V2 = 6.48 L, n1 = 0.225 mol
n2, and added moles
since n and V are directly proportional, when the volume increases, the moles should increase, and it does
By combing the gas laws we can write a general equation
R is called the gas constant
the value of R depends on the units of P and V
we will use 0.08206 and convert P to atm and V to L
the other gas laws are found in the ideal gas law if
two variables are kept constant
allows us to find one of the variables if we know the other 3
5.4 Ideal Gas Law
1 atm = 14.7 psi
T(K) = t(°C) + 273.15
P, V, T, R
How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C?
V = 3.24 L, P = 24.3 psi, t = 25 °C,
1 mole at STP occupies 22.4 L, since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas
Worked Example 5.3
since the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions
standard pressure = 1 atm
standard temperature = 273 K = 0°C
One mole of a gas occupy 22.41 L at STP
Calculate the (volume in liters occupied by 7.40g of NH3
n NH3 = 7.4 / 17 = 0.435 mol
V = nRT/P
V = 0.435 (0.0821) 273/1 = 9.74 l
Combined Gas Law
When we introduced Boyle’s, Charles’s, and Gay-Lussac’s laws, we assumed that one of the variables remained constant.
Experimentally, all three (temperature, pressure, and volume) usually change.
By combining all three laws, we obtain the combined gas law:
density is directly proportional to molar mass
1 atm = 760 mmHg, MM = 28.01 g
P, MM, T, R
Calculate the density of N2 at 125°C and 755 mmHg
P = 755 mmHg, t = 125 °C,
since the density of N2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is
Molar Mass of a Gas
From number of moles calculations
n = mass / M
PV= (mass / M ) RT
From density calculations
M = dRT/ P
1 atm = 760 mmHg,
Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg
m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,
molar mass, g/mol
m=0.311g, V=0.225 L, P=1.1658 atm, T=328 K,
the value 31.9 g/mol is reasonable
A chemist synthesized a greensh-yellow gaseous compound of chlorine and oxygen and find that its denity is 7.7g/L at 36°C and 2.88 atm. Calculate the molar mass and determine its molecular formula.
Molar mass = dRT/ P
ℳ = 7.7g/L ×0.0821×(36+273)/2.88 = 67.9 g/mol
Mass of empirical formula (ClO)= 35.45+16= 51.45
Ratio = Molar mass / Mass of empirical formula = 67.9/51.45= 1.3
molecular formula. ClO2
Calculate the volume of O2(in L) requred for the complete combustion of 7.64 L of (C2H2) measured at the same T & P
2 C2H2 (g) + 5O2 (g) → 4CO2 (g) + 2H2O ( ι)
From Avogadro low v= Rn
Volume of O2 = 7.64 L × 5L O2 /2L C2H2 = 19.1 L
2NaN3 (S) → 2 Na (s) + 3N2 (g)
Calculate the volume of N2 generate at 80°C and 823 mmHg by the decomposition of 60 g of NaN3
n of N2 = (60/65.02) × 3/2= 1.38
PV=nRT → V=nRT/P
V= 1.38 × 0.0821 ×(80+273)/ (823/760)
= 36.9 L
Dalton’s Law of Partial Pressures
V and T are constant
P total = PA + PB
nA + nB
mole fraction (Xi) =
Consider a case in which two gases, A and B, are in a container of volume V.
nA is the number of moles of A
nB is the number of moles of B
PT = PA + PB
PA = XA PT
PB = XB PT
Pi = Xi PT
Collecting a Gas Over Water
We can measure the volume of a gas by displacement.
By collecting the gas in a graduated cylinder, we can measure the amount of gas produced.
The gas collected is referred to as “wet” gas since it also contains water vapor.
PT = P O2 + P H2O
e value of R depends on the units of P and V
V= 1.38 × 0.0821 ×(80+273)/