The Ideal Gas Law
•
Ideal gas law combines the laws of Charles, Boyle,
and Avogadro
PV = nRT
•
1 mole of an ideal gas occupies 22.414 L at STP
•
STP - (P = 1atm, T = 273K (0°C))
•
R = 0.08206 Latm/moleK
01
Example
02
Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas.
Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel
vessel of volume 5.43 L at 69.5°C.
PV = nRT
P = nRT/V =
(1.82moles)(0.08206Lat/molK)(342.5K) = 9.42atm
(5.43L)
What is the volume (in liters) occupied by 7.40 g of CO2 at STP?
PV = nRT
V = nRT/P = (0.168moles)(0.08206Latm/molK)(273K) = 3.76L
(1atm)
7.40g x 1mole/44g = 0.168moles CO2
The Ideal Gas Law
Density and Molar Mass Calculations:
mass
nM PM
d


volume
V
R T
PV = nRT
n = PV/RT
03
Example
04
What is the molar mass of a gas with a density of
1.342 g/L at STP?
PV = nRT
n = PV/RT =
(1atm)(1L)
= 0.0446moles gas
(0.08206Latm/moleK)(273K)
1.342g/0.0446moles = 30.1g/mol = 30.1 amu
What is the density of uranium hexafluoride, UF6,
(MM = 352 g/mol) under conditions of STP?
PV = nRT
n = PV/RT =
(1atm)(1L)
(0.08206Latm/moleK)(273K)
352g/mol x 0.0446moles = 15.70g/L
= 0.0446moles gas
Example
05
The density of a gaseous compound is 3.38 g/L at
40°C and 1.97 atm. What is its molar mass?
PV = nRT
n = PV/RT = (1.97atm)(1L) = 0.0767 moles
(0.08206)(313K)
3.38g/0.0767mol = 44.07g/mol
Gas Stoichiometry
In gas stoichiometry at STP:
volume  moles
At STP:
1L air
1L Xe
1L O2
x moles
06
Gas Stoichiometry - Airbag movie 07
2NaN3(s) + (O)(s)
Na2O(s) + 3N2(g)
(O) is an oxidant (undergoes reduction-gains electrons)
145g of NaN3. How many liters of N2 gas are produced at 30°C and
1.15atm?
145g x 1mole/65.0g = 2.23moles NaN3
2.23moles NaN3 x 3moles N2
= 3.35moles N2
2moles NaN3
At STP  volume  moles
VN2 = 3.35moles x 22.4L/1mole = 75.04 L
NOT at STP!
V = nRT/P = (3.35mol)(0.08206Latm/molK)(303K) = 72.4L
(1.15atm)
Example
08
Hydrogen gas, H2, can be prepared by letting zinc metal react with
aqueous HCl. How many liters of H2 can be prepared at 742 mm Hg
and 15oC if 25.5 g of zinc (MM = 65.4 g/mol) was allowed to react?
Zn(s) + 2 HCl(aq)  H2(g) + ZnCl2(aq)
25.5g Zn x 1mole Zn x 1mole H2 = 0.390moles H2
65.39g Zn
1mole Zn
V = nRT/P = (0.390mol)(0.08206Latm/molK)(288K) = 9.44L H2
742mm Hg x 1atm/760mm Hg
Dalton’s Law of Partial Pressures 09
Air  N2 + O2 + Ar + CO2
Ptotal (air) = PN2 + PO2 + PAr + PCO2
PN2 = partial pressure
(pressure alone in container)
In a mixture of gases (Air) the total pressure (Ptotal) is the sum of the
partial pressures of the gases:
RT
P total 
V
n
Dalton’s Law of Partial Pressures 10
Partial pressure(N2) in air?
RT
P total 
V
Concentration of each component:
mole fraction (X) =
n
moles of component
total moles of the mixture
XN2 =
nN2
= nN2
nN2 + nO2 + nAr + nCO2
n = PV/RT
ntotal
XN2 = PN2 (V/RT) = PN2
Ptotal (V/RT)
Ptotal
PN2 = XN2 Ptotal
Example
11
Exactly 2.0 moles of Ne and 3.0 moles of Ar were placed
in a 40.0 L container at 25oC. What are the partial
pressures of each gas and the total pressure?
ntotal = 2moles + 3moles = 5moles
Ptotal = ntotalRT/V = (5moles)(0.08206Latm/molK)(298K) = 3.06atm
40.0L
PNe = XNe Ptotal = (2moles / 5moles)(3.06atm) = 1.224atm
PAr = XAr Ptotal = (3moles / 5moles)(3.06atm) = 1.836atm
Example
12
A sample of natural gas contains 6.25 moles of methane (CH4),
0.500 moles of ethane (C2H6), and 0.100 moles of propane (C3H8). If
the total pressure of the gas is 1.50 atm, what are the partial
pressures of the gases?
ntotal = 6.25mol + 0.500mol + 0.100mol = 6.85moles
PCH4 = XCH4 Ptotal = (6.25/6.85)(1.5atm) = 1.369
PC2H6 = XC2H6 Ptotal = (0.500/6.85)(1.5atm) = 0.109
PC3H8 = XC3H8 Ptotal = (0.100/6.85)(1.5atm) = 0.022
Ptotal = 1.369 + 0.109 + 0.022 = 1.5atm
Example
13
Hydrogen gas generated when calcium metal reacts with water is
collected at 30°C and a pressure of 988 mm Hg. The volume collected is
641 ml. What is the mass (in grams) of the hydrogen gas obtained? The
pressure of the water vapor at 30°C is 31.82 mm Hg.
Ca(s) + H2O(l)
CaO(s) + H2(g)
Ptotal = PH2 + PH2O PH2 = Ptotal - PH2O = 988mm Hg - 31.82mm Hg = 956.18mm Hg
n = PV/RT = (956.18mm Hg x 1atm/760mm Hg)(0.641L) = 0.0324 moles H2
(0.08206Latm/molK)(303K)
0.0324moles x 2g/1mole = 0.0648g = 64.8mg H2
Ideal versus nonideal gases
14
Some substances are ideal gases, while other
substances are nonideal gases  FALSE!
All gases are ideal under certain conditions (STP)
and nonideal under other conditions.
Ideal gases
15
Assumptions:
•
Volumes of the particles themselves are negligible
compared with the total volume of the gas
•
There are no attractive or repulsive forces between
particles
Behavior of Real Gases
16
Volume of the particles are not negligible!
The volume taken up by gas particles is more important at
higher pressures than at lower pressures. As a result, the
volume at high pressure will be greater than the ideal value.
Behavior of Real Gases
17
At higher pressures, particles are much closer together and
attractive forces become more important than at lower
pressures.
Molecules are drawn closer together, decreasing the volume.
Ideal gas law will over shoot the volume of the gas.