Gas Laws
Guided Notes
Mathematical Relationships
There is a Mathematical Relationship between
Pressure, Temperature and Volume of a constant
amount of gas
• Gas volumes change significantly with small
changes in temperature and pressure
• These changes can be defined by equations called
the gas laws.
• Gas laws are only valid for ideal gasses
• Ideal gases: do not exist, but are a model, they
have no attractive force and no volume
Gas Pressure
• Gas Pressure is related to the mass of the gas and
to the motion of the gas particles
• Gas molecules move and bounce off the walls of
their container
– These collisions cause gas pressure
• Pressure is a force per unit area
– Standard Units of pressure = Pascal(Pa)
– 1 Pa is the pressure of 1 Newton per square meter
(N/m2)
• Normal air pressure at sea level is 1 atmosphere (atm)
• 1 kPa=1000Pa
– Units of pressure: atmosphere, mm Hg, kPa, psi
• 1.00 atm = 760 mm Hg = 101.325 kPa =14.7 psi
Atmospheric Pressure
• The pressure exerted on the Earth by the gasses in the
atmosphere
• Absolute pressure will include the pressure on a closed
system, as indicated by a gauge PLUS the pressure
exerted by the atmosphere.
Ex.
The pressure gauge on a bicycle tire reads 44 psi, what is
the absolute pressure?
44psi+14.7psi =59 psi
pressure reading + atmospheric pressure = absolute pressure
To Convert to kPa;
59 psi X (101.3kPa/14.7 psi) =410kPa
Temperature Scales
• In order to use all of the gas laws, you must
express the temperature in degrees kelvin (K)
– To convert from celsius (°C) to kelvin (K), use the
following conversion factor:
K= 273 + °C
Boyle’s Law
Relates Pressure and Volume
• In an ideal gas, with constant temperature, if
the pressure increases, the volume decreases.
Boyle’s Law: GAS VOLUME AND PRESSURE, at
constant temperature, ARE INVERSELY
PROPORTIONAL! ( )
Boyle’s Law Example Problem
1. If 425 mL of O2 are collected at a pressure of 9.80 kPa
what volume will the gas occupy if the pressure is
changed to 9.40 kPa?
– To find the new volume, you need the original volume AND the
change in pressure
𝑃1
𝑉2 = 𝑉1 ×
𝑃2
The pressure decreases from 9.8 to 9.4 kPa. (V will increase)
𝑃1
𝑉2 = 𝑉1 ×
𝑃2
425𝑚𝐿 9.8 k𝑃𝑎
𝑉2 =
×
= 443.0 mL
1
9.4 𝑘𝑃𝑎
Boyle’s Law Practice Problem
2. Calculate the pressure of a gas that occupies a
volume of 125.0 mL, if at a pressure of 95.0 kPa,
it occupies a volume of 219.0 mL.
–
To find the new pressure, you need the original pressure AND the
change in volume
𝑉1
𝑃2 = 𝑃1 ×
𝑉2
The volume decreases from 219.0 mL to 125.0mL.(P will increase)
𝑉1
𝑃2 = 𝑃1 ×
𝑉2
95.0 𝑘𝑃𝑎 219.0 𝑚𝐿
𝑃2 =
×
= 166.0 kPa
1
125.0 𝑚𝐿
Charles’s Law
Relates Temperature and Volume
• In an ideal gas, with constant pressure, if the
temperature increases the volume increases.
Charles’s Law: GAS VOLUME AND KELVIN
TEMPERATURE, at constant pressure, are
DIRECTLY PROPORTIONAL.
You must express temp in degrees K
Charles’s Law Example Problem
1.
What volume will a sample of nitrogen occupy at 28.0 degrees C if
the gas occupies a volume of 457 mL at a temperature of 0.0
degrees C? Assume the pressure remains constant.
First: Convert temp to K
T in K= 273 + °C
T2 = 273 + 28.0 °C = 301 K
and T1 = 273 + 0.0 °C = 273 K
Second: To find the new volume, you need the original volume AND
the change in Kelvin temperature
𝑉2 = 𝑉1 ×
𝑇2
𝑇1
The temperature increases from 273 K to 301 K. (V will increase)
(If volume increases, should be multiplied by a ratio > than 1)
𝑉2 =
457 𝑚𝐿 301 𝐾
×
= 504 mL
1
273 𝐾
Charles’s Law Practice Problem
2.
If a gas occupies a volume of 733 mL at 10.0 °C, at what temperature in
°C, will it occupy a volume of 1225 mL if the pressure remains constant?
First: Convert temp to K
Convert °C to °K (°K = °C + 273) 10.0 °C = 283 °K
Second: To find the new Kelvin temperature, you need the original Kelvin
temperature and the change in volume.
𝑉2
𝑇2 = 𝑇1 ×
𝑉1
The volume increases from 733 mL to 1225 mL. (T will increase)
(If temperature increases, should be multiplied by a ratio > than 1)
𝑇2 =
283 𝐾 1225 𝑚𝐿
×
= 473 K
1
733 𝑚𝐿
Last: Convert temp from K to °C
°C = 473 K – 273 = 200 °C
Gay- Lussac’s Law
Relates Temperature and Pressure
• In an ideal gas, with constant volume, if the
pressure increases the Kelvin temperature
increases.
Gay-Lussac’s Law: GAS PRESSURE AND KELVIN
TEMPERATURE, at constant volume, are
DIRECTLY PROPORTIONAL.
You must express temp in degrees K
Gay- Lussac’s Law Example Problem
1.
A cylinder of gas has a pressure of 4.40 atm at 25 °C. At what temperature, in Celsius, will it reach
a pressure of 6.50 atm?
First: Convert temp to K
T1 = 25 °C +273=298 K
Second: To find the new temperature, you need the original temperature AND the change in pressure
𝑃1
𝑇1
=
𝑃2
𝑇2
OR
𝑃1 𝑇2 = 𝑃2 𝑇1
The pressure increases from 4.40atm to 6.50atm. (T will increase)
4.40 𝑎𝑡𝑚 6.50 𝑎𝑡𝑚
=
298 𝐾
𝑇2
4.40 𝑎𝑡𝑚 × 𝑇2 = 6.50 𝑎𝑡𝑚 × 298 𝐾
4.40 𝑎𝑡𝑚 × 𝑇2 6.50 𝑎𝑡𝑚 × 298 𝐾
=
4.40 𝑎𝑡𝑚
4.40 𝑎𝑡𝑚
𝑇2 = 440 𝐾
Last: Convert temp from K to °C
°C = 440 K – 273 = 167 °C
Gay- Lussac’s Law Practice Problem
1.
A cylinder of gas has a pressure of 3,350 psi at 2 °C. What will the pressure be at 24 °C?
First: Convert temp to K
T1 = 2 °C + 273 = 275 K
T2 = 24 °C + 273 = 297 K
Second: To find the new pressure, you need the original pressure AND the change in temperature
𝑃1
𝑇1
=
𝑃2
𝑇2
OR
𝑃1 𝑇2 = 𝑃2 𝑇1
The temperature increases from 275 K to 297 K. (P will increase)
3350 𝑝𝑠𝑖
𝑃2
=
275 𝐾
297 𝐾
3350 𝑝𝑠𝑖 × 297 𝐾 = 𝑃2 × 275 𝐾
3350 𝑝𝑠𝑖 × 297 𝐾 𝑃2 × 275 𝐾
=
275 𝐾
275 𝐾
𝑷𝟐 = 𝟑𝟔𝟏𝟖 𝒑𝒔𝒊
Combined Gas Law
Changing Temperature and Pressure
• Boyle’s and Charles’s laws together make up the
Combined Gas Law.
• A pressure ratio and a Kelvin temperature ratio are
needed to calculate the new volume.
𝑃1 𝑇2
𝑉2 = 𝑉1 ∙ ∙
𝑃2 𝑇1
• STP (Standard Temperature and Pressure) = 273K and
101.3 kPa
Combined Gas Law Example Problem
1. Calculate the volume of a gas at STP if 502 mL of the gas
are collected at 29.7 °C and 96.0 kPa
First: Convert Celsius to Kelvin temps. (29.7°C + 273 = 302.7 K)
Second: Organize the given data in a table.
Value
Initial Condition
New Conditions
Gas Volume Change?
P
96.0 kPa
101.3 kPa
Decrease
V
502 mL
?
?
T
302.7 K
273 K
Decrease
𝑉2 = 𝑉1 ∙
Calculate:
96.0 𝑘𝑃𝑎 273 𝐾
𝑉2 = 502 𝑚𝐿 ∙
∙
101.3 𝑘𝑃𝑎 302.7 𝐾
𝑽𝟐 = 𝟒𝟐𝟗 𝒎𝑳
𝑃1
𝑃2
∙
𝑇2
𝑇1
Combined Gas Law Practice Problem
If 400 ml of oxygen are collected at 20.0C, and the
atmospheric pressure is 94.7 kPa, what is the volume of the
oxygen at STP?
Value
Initial Condition
New Conditions
Gas Volume Change?
P
94.7 kPa
101.3 kPa
Decrease
V
400. mL
?
?
T
293 K
273 K
Decrease
94.7 𝑘𝑃𝑎 273 𝐾
𝑉2 = 400 𝑚𝐿 ∙
∙
101.3 𝑘𝑃𝑎 293 𝐾
𝑽𝟐 = 𝟑𝟒𝟖 𝒎𝑳
Avogadro’s Law
Relates Volume and Moles
• For a gas at constant temperature and
pressure, the volume is directly proportional
to the number of moles of gas.
• Relates the quantity of gas particles to the
volume.
𝑉1 × 𝑛2 = 𝑉2 × 𝑛1
Progression of Laws
• Boyles', Charles', and Avogadro's laws
combine to form the ideal gas law, which is
the über law of gases.
• The ideal gas law can be manipulated to
explain Dalton's law of partial pressures, gas
density, and the mole fraction. It can also be
used to derive the other gas laws.
Ideal Gas Law
• The ideal gas law is an ideal law. It operates
under a number of assumptions.
• Two important assumptions:
– the molecules of an ideal gas do not occupy space
– the molecules of an ideal gas do not attract each
other
• These assumptions work well at the relatively low
pressures and high temperatures, but there are
circumstances in the real world for which the
ideal gas law holds little value.
Ideal Gas Law
𝑷∙𝑽=𝒏∙𝑹∙𝑻
•
•
•
•
P is pressure in kPa
V is volume in L
T is temperature in K
n is number of moles
• R is a constant = 8.31
𝑳∙𝒌𝑷𝒂
𝒎𝒐𝒍∙𝑲
– R can have different units when needed
• This equation can be used to determine the molecular mass of a gas
𝑛=
𝑚
𝑀
𝑃∙𝑉 =
𝑚
∙𝑅∙𝑇
𝑀
𝑤ℎ𝑒𝑟𝑒 𝒏 𝑖𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠, 𝒎 𝑖𝑠 𝑚𝑎𝑠𝑠 𝑎𝑛𝑑 𝑴 𝑖𝑠 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
Ideal Gas Law Example Problem
1. How many moles of gas will a 1250 mL flask
hold at 35.0 degrees C and a pressure of 95.4
kPa?
𝑃𝑉 = 𝑛𝑅𝑇
𝑜𝑟
𝑛=
First: Convert temp to Kelvin
T = 273+ 35.0°C = 308K
𝑛=
𝑃𝑉
𝑅𝑇
=
95.4 𝑘𝑃𝑎×1250 𝑚𝐿
𝑃𝑉
𝑅𝑇
1𝐿
×
𝐿∙𝑘𝑃𝑎
1000 𝑚𝐿
8.31
×308 𝐾
𝑚𝑜𝑙∙𝐾
𝒏 = 𝟎. 𝟎𝟒𝟔𝟔 𝒎𝒐𝒍
NOTE:
mL must be
converted to L to
match units in the
constant
1L = 1000 mL
Ideal Gas Law Practice Problem
2. A flask has a volume of 258 mL. A gas with mass 1.475 g
is introduced into the flask at a temperature of 302.0 K
and a pressure of 9.86 x104 Pa. Calculate the molecular
mass of the gas using the ideal gas equation.
𝑛=
𝑀=
𝑚
𝑀
𝑀=
1.475 𝑔 × 9.86×104 𝑃𝑎 ×258 𝑚𝐿
𝐿∙𝑘𝑃𝑎
8.31
×302.0 𝐾
𝑚𝑜𝑙∙𝐾
𝑚𝑃𝑉
𝑅𝑇
×
𝒈
𝑴 = 𝟏𝟒𝟔. 𝟎
𝒎𝒐𝒍
1𝑘𝑃𝑎
1000 𝑃𝑎
×
1𝐿
1000 𝑚𝐿
Compare and Contrast Gas Laws
Gas Law
Relates
Equation
Unit
Boyle’s
Pressure to Volume
P1V1=P2V2
L or kPa
Charles’s
Temperature to
Volume
T1V2=T2V1
K or L
Combined Gas
law
Temperature,
pressure and volume
V2 =V1(P1/P2)(T2/T1)
K, L and
kPa
Gay- Lussac’s Law
Temperature and
pressure
P1T2=P2T1
K or kPa
Avogadro’s Law
volume to moles
V1 / n1 = V2 / n2
kPa or
mol
Ideal Gas Law
Pressure, volume,
temp and moles
PV=nRT and
PV = m RT
M
Mol, L, K,
or kPa
Ideal Gas Law Problems
1. How many moles of He are contained in a
5.00L canister at 101 kPa and 30.0 dC?
2. What is the volume of 0.020 mol Ne at 0.505
kPa and 27.0 dC?
3. How much Zn must react in order to form
15.5 L of H2 gas at 32.0 dC and 115 kPa?
Zn + H2SO4  ZnSO4 + H2
Ideal Gas Law Problem Solutions
1. How many moles of He are contained in a 5.00L
canister at 101 kPa and 30.0 dC?
n= VP/RT
(5.00 L) (101 kPa)
( 8.31 kPa L)
303 K
(1 mol K)
= 5.0 X 101 X 1 mol
8.31 X 303
= 0.201 mol
Ideal Gas Law Problem Solutions
2. V= nRT
P
V = ( 0.020 mol Ne) (8.31 kPa L) (300 K)
( 1 mol K)
0.505 kPa
V= 99 L Ne
Ideal Gas Law Problem Solutions
3. First determine moles
PV=nRT n=PV
n= (115 kPa) ( 15.5 L)
RT
( 8.31 kPa L) (305)
( 1 mol K )
n= 0.703 mol H2
Now determine mass of Zn (molar ratio and molar
mass)
(0.703 mol H2) (1 mol Zn) ( 65.39 g Zn)
(1)
(1 mol H2) ( 1 mol Zn)
= 46.0 g Zn