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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
● 1 Liter = 1 decimeter
● 1 cubic centimeter (cc) = 1 mL
● 1 L = 1000 mL = 1000 cc
3. Temperature Scales
● ° = ξͺξšξ™ξ·Β° βˆ’ 
● K = CΒ° + 273.15
● ξ²ξŽ“ξŽ€ξŽ ξŽξŽ¦ξŽ₯ξŽ– ξŽ«ξŽ–ξŽ£ξŽ   ξŽ₯ξŽ–ξŽžξŽ‘ξŽ–ξŽ£ξŽ’ξŽ₯ξŽ¦ξŽ£ξŽ–   ξΌξŽ–ξŽξŽ§ξŽšξŽŸ  βˆ’ξ£ξ¨ξ€ξŸξ’ξ¦
4. Pressure
● Measured as force acting on an area
● P = (F / A)
β—‹ F = force
β—‹ A = area
β—‹ In pounds / in2
● ξŽ„ξŽ”ξŽšξŽ–ξŽŸξŽ₯ξŽšξŽ—ξŽšξŽ” ξ™ξŽœξŽ˜ βˆ’ ξ™ξŽž  2)) / m2 which is a Pascal (Pa) *don’t need to
know
● We primarily use relative measurements of pressures, since we are always
surrounded by gases
● 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
● 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. ξŽξŽ£ξŽ–ξŽ€ξŽ€ξŽ¦ξŽ£ξŽ– ξŽ’ξŽŸξŽ• VξŽ ξŽξŽ¦ξŽžξŽ– ξŽƒξŽ–ξŽξŽ’ξŽ₯ βˆ’ ξŽͺξŽξŽ–β€™ξŽ€ ξ½ξŽ’ξŽ¨,  
● As volume decreases, pressure increases
β—‹ 1 L at 1 atm
β—‹ 0.5 L at 2 atm
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● Two Expressions
β—‹ V ∝ 
β—‹ PV = constant
2. VξŽ ξŽξŽ¦ξŽžξŽ– ξŽ’ξŽŸξŽ• ξŽ…ξŽ–ξŽžξŽ‘ξŽ–ξŽ£ξŽ’ξŽ₯ξŽ¦ξŽ£ξŽ– ξŽƒξŽ–ξŽξŽ’ξŽ₯ βˆ’ ξ΄ξŽ™ξŽ’ξŽ£ξŽξŽ–ξŽ€β€™ ξ½ξŽ’ξŽ¨
● V ∝ ξŽ…
● T is in Kelvin
β—‹ 10Β°C β†’ 20Β°C ***no change
β—‹ Above statement is equivalent to 283 K β†’ 293 K
● At constant pressure the volume of a given quantity of gas is proportional
to absolute temperature.
3. ξŽ‚ξŽ¦ξŽ’ξŽŸξŽ₯ξŽ₯ξŽͺ ξŽ’ξŽŸξŽ• VξŽ ξŽξŽ¦ξŽžξŽ– ξŽƒξŽ–ξŽξŽ’ξŽ₯ βˆ’ ξ²ξŽ§ξŽ ξŽ˜ξŽ’ξŽ•ξŽ£ξŽ β€™ξŽ€ ξ½ξŽ’ξŽ¨
● Gay-Lussac’s combining volumes
● H2 + Cl2 β†’ 2HCl
● 1 volume + 1 volume β†’ 2 volumes
● Equal volumes at the same pressure and temperature contain equal
numbers of molecules.
● V = k N
β—‹ N is number of molecules
β—‹ V Ξ± N
β—‹ V Ξ± n
● Molar Volume
β—‹ 1 mole O2 βˆ’   ξŽ£ξŽ–ξŽ’ξŽ¦ξŽšξŽ£ξŽ–ξŽ€ ξ₯  ξŽ’ξŽ₯ ξŽ„ξŽ…ξŽ
β—‹ 1 mole H2 βˆ’   ξŽ£ξŽ–ξŽ’ξŽ¦ξŽšξŽ£ξŽ–ξŽ€ ξ₯  ξŽ’ξŽ₯ ξŽ„ξŽ…ξŽ
β—‹ at STP (273.15 K and 1 atm) 1 mole of any gas occupies 22.4 L
C. Ideal Gas Equations
● V Ξ± 1 / P
● V Ξ± T
● V Ξ± (nT) / P
● V Ξ± n
● 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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● P = 1.00 atm
● V = 22.4 L
● n = 1.00 mol
● 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)
● n = (18 g H2O) / (1 mol H2O / 18 g H2O) = 1 mol
● R = 0.0821
● T = 100Β°C + 273 = 373 K
● 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)
● n = 3.00 mol
● R = 0.0821
● T = 100Β°C + 273 = 373 K
● 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
● Boyle’s Law
● PV = nRT = constant
● P1V1 = P2V2
b. For a given quantity of gas at constant pressure
● Charles’ Law
● PV = nRT
● V / T = nR / P = constant
● V1 / T1 = V2 / T2
c. For a given quantity of gas
● PV = nRT
● (PV) / (T) = nR = constant
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