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Chapter 10

# CHE 101 Chapter Notes - Chapter 10: Van Der Waals Equation, Molar Mass, Gas Laws

Department
Chemistry
Course Code
CHE 101
Professor
Atwood Jim
Chapter
10

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Chapter 10 - Gases
Jim Atwood
A. Definitions
1. Gas - Molecules of a ga are not held together, expand to fill the space available
and can be compressed
2. Volume is the space occupied
β 1 Liter = 1 decimeter
β 1 cubic centimeter (cc) = 1 mL
β 1 L = 1000 mL = 1000 cc
3. Temperature Scales
β ξ΄Β° = ξξ¦ξ ξͺξξξ·Β° β ξ€ξ£ξ
β K = CΒ° + 273.15
β ξ²ξξ€ξ ξξ¦ξ₯ξ ξ«ξξ£ξ  ξξ ξ₯ξξξ‘ξξ£ξξ₯ξ¦ξ£ξ ξξ€ ξ‘ ξΌξξξ§ξξ ξ ξ£ βξ£ξ¨ξ€ξξ’ξ¦
4. Pressure
β Measured as force acting on an area
β P = (F / A)
β F = force
β A = area
β In pounds / in2
β ξξξξξξ₯ξξξξ ξξξ β ξξ ξ  ξ€2)) / m2 which is a Pascal (Pa) *donβt need to
know
β We primarily use relative measurements of pressures, since we are always
surrounded by gases
β A barometer at sea level will measure 760 mm of Hg - defined as 1 atm
β 1 atm = 760 mm Hg
β 1 mm Hg = 1 Tor
β STP (standard temperature and pressure) 1 atm (760 mm Hg) and 0Β°C
(273 K)
Example: Convert 0.605 atm into pressure in Torr.
(0.605 atm)(760 Torr / 1 atm) = 460 Torr
B. Gas Laws
1. ξξ£ξξ€ξ€ξ¦ξ£ξ ξξξ Vξ ξξ¦ξξ ξξξξξ₯ξξ ξξ€ β ξ³ξ ξͺξξβξ€ ξ½ξξ¨, ξ·ξξ ξ’ξ‘ξξ¨
β As volume decreases, pressure increases
β 1 L at 1 atm
β 0.5 L at 2 atm
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β Two Expressions
β V β ξ’ξ ξ
β PV = constant
2. Vξ ξξ¦ξξ ξξξ ξξξξ‘ξξ£ξξ₯ξ¦ξ£ξ ξξξξξ₯ξξ ξξ€ β ξ΄ξξξ£ξξξ€β ξ½ξξ¨
β V β ξ
β T is in Kelvin
β 10Β°C β 20Β°C ***no change
β Above statement is equivalent to 283 K β 293 K
β At constant pressure the volume of a given quantity of gas is proportional
to absolute temperature.
3. ξξ¦ξξξ₯ξξ₯ξͺ ξξξ Vξ ξξ¦ξξ ξξξξξ₯ξξ ξξ€ β ξ²ξ§ξ ξξξξ£ξ βξ€ ξ½ξξ¨
β Gay-Lussacβs combining volumes
β H2 + Cl2 β 2HCl
β 1 volume + 1 volume β 2 volumes
β Equal volumes at the same pressure and temperature contain equal
numbers of molecules.
β V = k N
β N is number of molecules
β V Ξ± N
β V Ξ± n
β Molar Volume
β 1 mole O2 β ξ€ξ£ ξ ξ£ξξ’ξ¦ξξ£ξξ€ ξ£ξ£ξξ₯ ξ½ ξξ₯ ξξξ
β 1 mole H2 β ξ£ ξ ξ£ξξ’ξ¦ξξ£ξξ€ ξ£ξ£ξξ₯ ξ½ ξξ₯ ξξξ
β at STP (273.15 K and 1 atm) 1 mole of any gas occupies 22.4 L
C. Ideal Gas Equations
β V Ξ± 1 / P
β V Ξ± T
β V Ξ± (nT) / P
β V Ξ± n
β V = (nRT) / P
β R is the gas constant
β The Ideal Gas Equation, which has all the variables that describes a gas
Example 1: Calculate the value of R
PV = nRT
R = (PV) / (nT)
R = ((1)(22.4)) / ((1)(273))
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β P = 1.00 atm
β V = 22.4 L
β n = 1.00 mol
β T = 0Β°C + 273 = 273 K
R = 0.821 ((L)(atm)) / ((mol)(K))
Example 2: Calculate the pressure exerted by 18 g of steam confined to 18 L at 100Β°C.
PV = nRT
P = (nRT) / V
P = ((1)(0.0821)(373)) / (18)
β n = (18 g H2O) / (1 mol H2O / 18 g H2O) = 1 mol
β R = 0.0821
β T = 100Β°C + 273 = 373 K
β V = 18 L
P = 1.7 atm
Example 3: What volume would be required to contain 3.00 moles of O2 at 100Β°C and
2.00 atm pressure?
PV = nRT
V = (nRT) / P
V = ((3.00)(0.821)(373)) / (2.00)
β n = 3.00 mol
β R = 0.0821
β T = 100Β°C + 273 = 373 K
β P = 2.00 atm
V = 45.9 L
You should be able to solve any problem given three of the four variables.
1. Relation to other gas laws:
a. For a given quantity of gas at constant temperature
β Boyleβs Law
β PV = nRT = constant
β P1V1 = P2V2
b. For a given quantity of gas at constant pressure
β Charlesβ Law
β PV = nRT
β V / T = nR / P = constant
β V1 / T1 = V2 / T2
c. For a given quantity of gas
β PV = nRT
β (PV) / (T) = nR = constant
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