Gas Laws
Q4U1 (13d)
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 101.325 kilopascals (kPa)
• 101.325 kPa = 760mm Hg, 1kPa = 7.5 mm Hg (1 kPa=1000Pa)
– Other units of pressure, Atmosphere or 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
Convert to kPa;
59 psi X (101.3kPa/14.7 psi) =410kPa
Barometer vs. Manometer: Both
measure pressure
Barometer: is always closed end
Calculations for Open Ended System
• If mmHg is higher in open end
• Pgas =Patmosphere + Ph3
Solve for gas pressure if:
Ph3 = 185 mmHg
Atmospheric pressure is 98.95 kPa
1. Convert mmHg to kPa
# kPa = (185mm/1)(1kPa/7.50mm)=24.7 kPa
2. Add kPa to atmospheric pressure
Pressure of gas = 98.95 +24.7= 124kPa
Closed system: Used to measure
pressure of a gas
•The closed arm is filled with gas, what is the pressure of the
gas, in kPa?
•The difference in the Hg level is 165 mm
•7.50 mm = 1 kPa
Pressure of gas = (165.0 mm/1) ( 1kPa/7.5 mm) = 22.0 kPa
Try some yourself
1. Closed manometer containing sulfur dioxide gas
(SO2). Height difference in mmHg is 560, what is
the pressure of SO2 in kPa?
2. An open monometer containing Hydrogen. Hg
level is 78.0 mmHg in the arm connected to the
air (open end), air pressure is 100.7kPa. What is
the pressure of Hydrogen in kPa?
3. An open monometer containing Nitrogen. Hg
level is 26.0 mmHg in the arm connected to the
gas, air pressure is 99.6kPa. What is the pressure
of nitrogen in kPa?
Answers
1. Pressure of SO2 = (560.0 mm/1) ( 1kPa/7.5
mm) = 74.6 kPa
2. # kPa = (78.0mm/1)(1kPa/7.50mm)=10.4 kPa
Pressure of H2= 100.7kPa + 10.4kPa= 111.1 kPa
3. # kPa= (26.0mm/1)(1kPa/7.50mm)= 3.47kPa
Pressure of N2 = -3.47kPa + 99.6 =96.1kPa
notice the negative number, less than air pressure
Homework
• Complete section review #1-8 handout p 73
Mathematic 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
Boyle’s Law; Relates Pressure and Volume
• If the pressure on an ideal gas is increased, the volume
decreases
• When the pressure is doubled ( ), the volume of the gas is half
( ) what it was
• If the pressure is cut( ) by half, the volume is doubled ( ).
Boyle’s Law: GAS VOLUME AND PRESSURE, at constant
temperature, ARE INVERSELY PROPORTIONAL! ( )
(when one goes up the other goes down)
– To find the new volume, you need the original volume AND the
change in pressure
V2 = V1 X P1/P2
– To find new pressure, you need the original pressure AND the
change in volume --- New pressure = old pressure X volume Ratio
P2 = P1 X V1/V2
- Mentally decide if the new volume is larger or smaller than the
old, arrange the pressure ratio accordingly (If larger use fraction
greater than 1, If smaller use fraction less than 1)
Boyle’s Law Example Problems
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?
• the pressure decreases from 9.8 to 9.4 kPa. (V will increase)
V2 = V1 X P1/P2
(425mL/1) ( 9.8 kPa/9.4 kPa) = 443.0 mL
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.
• the volume decreases from 219.0 mL to 125.0mL.(P will
increase)
P2 = P1 X V1/V2
(95.0 kPa/1) (219.0 mL/125.0mL) = 166.0 kPa
Charles’s Law; relates Temperature and
Volume
• In an ideal gas, with constant pressure, if the
temperature increases the volume will increase.
• Charles’s Law: at a constant pressure, the volume of a
gas is directly proportional to its Kelvin temperature.
(you must express temp in degrees K)
• New volume =old volume X degree K temp change
V2 = V1 X T2/T1
• New temperature = old temp (K) X volume change
T2 = T1 X V2/V1
- Mentally decide if the new volume is larger or smaller than
the old, arrange the pressure ratio accordingly (If larger use
fraction greater than 1, If smaller use fraction less than 1)
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.
convert temp to K
K= 273 + degrees C
K = 273 + 28.0 degrees C = 301 K
and K = 273 + 0.0 C = 273 K
(Temp increase means Volume increase: Ratio > than 1)
New Volume = (457mL/1) (301 k/273 k) = 504 mL
Charles’s Law Example Problem #2
2. If a gas occupies a volume of 733 mL at 10.0 dC, at
what temperature in C degrees, will it occupy a
volume of 1225 mL if the pressure remains
constant?
(increased volume means increased temperature)
Convert °C to °K (°K = °C + 273) 10.0 °C = 283 °K
T2 = T1 X V2/V1
= (283 °K/1) (1225 mL/733 mL) = 473 °K
Convert °K to °C
°C = 473 K – 273 = 200 °C
Practice use of Boyle’s Law
• Complete the problem solving packet, Pg 1216; problems 1-5 all parts
• Show all work
• Label all units
• Due tomorrow (complete for homework if
necessary)
Practice use of Charles’s Law
• Complete the problem solving packet, Pg 17-20
problems 1-3 all parts
• Show all work
• Label all units
• Due tomorrow (complete for homework if
necessary)
Combined Gas Law: when both
temperature and pressure change occur
• Boyle’s and Charles’s laws used together make the
combined gas law.
• A pressure ratio and a Kelvin temperature ratio are
needed to calculate the new volume.
New Volume = old volume X pressure ratio X Kelvin temperature ratio
V2 =V1(P1/P2)(T2/T1) *STP=273K and 101.3kPa
• First: Construct a press-volume-temp data table
Value
P
V
T
Old Conditions
New Conditions
What happens to gas
Volume?
Practice: Combined Gas Law
Calculate the volume of a gas at STP if 502 mL of the gas are
collected at 29.7 C and 96.0 kPa
•Convert Celsius to Kelvin temps. ( 29.7 + 273 = 302.7)
•Organize the data in the table.
Value
Old Conditions
New Conditions
What happens to gas
Volume?
P
V
T
96.0 kPa
502 mL
302.7 K
101.3 kPa
?
273 K
Decrease
?
Decrease
Calculate:
V2 =V1(P1/P2)(T2/T1)
V2 = 502 mL ( 96.0kPa/ 101.3kPa) (273K/ 302.7k)
V2 = 429 mL
Try this on your own!
• 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
Old Conditions
New Conditions
What happens to gas
Volume?
P
V
T
94.7 kPa
400. mL
293 K
101.3 kPa
?
273 K
Decrease
?
Decrease
V2 =V1(P1/P2)(T2/T1)
= 400 mL (94.7 kPa/101.3 kPa) (273 k/293 K)
= 348 mL
Homework: Complete Handout, Pg 78-79
•
•
•
•
#’s 10-14, all parts
Show all of your work
label all of your units
DUE TOMORROW!!!
Gay- Lussac’s Law: relates Temperature
and pressure
• The Pressure of a gas is DIRECTLY proportional
to the absolute temperature when the volume
is unchanged.
• P1T2=P2T1 or P1 = P2
T1 T2
• When temp increases, pressure increases
• Must use Kelvin temperatures!
Gay- Lussac’s Law Example
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?
P1 = 4.40atm
P1T2=P2T1
T2= ? K and ?C
298 X 6.50 atm = 440 K
P2 = 6.50 atm
4.40 atm
T1 = 25+273=298 K 440.K = (440. – 273) C =167°C
Practice Problems; Gay- Lussac’s Law
• Complete the problem solving packet, Pg2024, problems 1-3 all parts
• Show all work
• Label all units
Avogadro’s Law; relates volume to
moles
• Avogadro's Law states: for a gas at constant
temperature and pressure, the volume is directly
proportional to the number of moles of gas.
• Avogadro's law relates the quantity of a gas and
its volume.
• According to Avogadro's Law: V1n2 = V2n1
• When any three of the four quantities in the
equation are known, the fourth can be calculated.
– For example, if n1, V1 and V2 are known, the n2 can be
solved by the following equation:
– n2 = V2 x (n1/V1)
Avogadro’s Law Practice
• Example #1: 5.00 L of a gas is known to
contain 0.965 mol. If the amount of gas is
increased to 1.80 mol, what new volume will
result (at an unchanged temperature and
pressure)?
• Answer: this time I'll use V1n2 = V2n1
• (5.00 L) (1.80 mol) = (x) (0.965 mol)
Avogadro’s Law Practice
Example #2: A cylinder with a movable piston contains 2.00 g of
helium, He, at room temperature. More helium was added to
the cylinder and the volume was adjusted so that the gas
pressure remained the same. How many grams of helium were
added to the cylinder if the volume was changed from 2.00 L to
2.70 L? (The temperature was held constant.)
Solution:
1) Convert grams of He to moles:
2.00 g / 4.00 g/mol = 0.500 mol
2) Use Avogadro's Law:
V1/n1 = V2/n2
2.00 L / 0.500 mol = 2.70 L / x
x = 0.675 mol
3) Compute grams of He added:
0.675 mol - 0.500 mol = 0.175 mol 0.175 mol x 4.00 g/mol =
0.7 grams of He added
Progression of Laws
• Boyles', Charles', and Avogadro's laws
combine to form the ideal gas law, which is
the uber law of gases.
• The ideal gas law can be manipulated to
explain Dalton's law, partial pressure, 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.
• The two most important assumptions are that
the molecules of an ideal gas do not occupy
space and 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
• Equation: PV = nRT
P is pressure in kPa
V is volume in cubic decimeters
T is temperature in K
n represents the number of moles of the gas
R is a constant , using these units, is 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
– Moles (n) = mass(m)/ molecular mass(M)
PV = m RT
M
Amount of Gas Present
• Amount of gas present is related to its
pressure, temperature and volume using the
Ideal Gas Law
• Applying the gas law allows for the calculation
of the amount of gaseous reactants and
products in reactions
Application Of Gas Laws Example
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 = PV
RT
Convert degree C to Kelvin, 35.0C = 273+ 35.0 = 308K
n = PV
RT
n= 95.4 kPa
1250 mL
1L
8.31 L kPa/mol K 308 K
1000 mL
n= 0.0466 mol
Gas Laws Example 2
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= mass/molecular mass(M)
M = mRT
PV
M= 1.475 g 8.31 L kPa 302.0 K 1000 mL 1000 Pa
9.86 x104 Pa mol K
258 mL
1L
1kPa
M= 146.0 g/mol
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
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 / n 1 = V 2 / n 2
kPa or
mol
Ideal gas Law
Pressure, volume,
temp and moles
PV=nRT and
PV = m RT
M
Mol, L, K,
or kPa