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 LawRelates 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

The pressure decreases from 9.8 to 9.4 kPa. (V will increase)

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

The volume decreases from 219.0 mL to 125.0mL.(P will increase)

Charles’s LawRelates 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

The temperature increases from 273 K to 301 K. (V will increase)

(If volume increases, should be multiplied by a ratio > than 1)

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 °KSecond: To find the new Kelvin temperature, you need the original Kelvin

temperature and the change in volume.

The volume increases from 733 mL to 1225 mL. (T will increase)(If temperature increases, should be multiplied by a ratio > than 1)

Last: Convert temp from K to °C °C = 473 K – 273 = 200 °C

Gay- Lussac’s LawRelates 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 Problem1. 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 pressureOR

The pressure increases from 4.40atm to 6.50atm. (T will increase)

Last: Convert temp from K to °C °C = 440 K – 273 = 167 °C

Gay- Lussac’s Law Practice Problem1. 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

OR

The temperature increases from 275 K to 297 K. (P will increase)

Combined Gas LawChanging 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.

• STP (Standard Temperature and Pressure) = 273K and 101.3 kPa

Combined Gas Law Example Problem

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

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.

Calculate:

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

Avogadro’s LawRelates 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.

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 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

– R can have different units when needed

• This equation can be used to determine the molecular mass of a gas

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?

First: Convert temp to Kelvin

T = 273+ 35.0°C = 308K

NOTE:mL must be converted to L to match units in the constant 1L = 1000 mL

Ideal Gas Law Practice Problem2. 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.

Compare and Contrast Gas LawsGas 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 H2Now 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