Section 3.7—Gas Laws

How can we calculate Pressure, Volume and Temperature of our airbag?

Objectives:•Compare different pressure units•Relate the Kelvin and Celsius scales for temperature•Define standard temperature and pressure•Define and perform calculations for 5 gas laws

Pressure UnitsPressure can be measured using several different units.

Unit Symbol

atmospheres atm

Pascals, kiloPascals

millimeters of mercury or torr

pounds per square inch

Pa, kPa

mm Hg, torr

psi

Pressure Units

It is fairly easy to convert from one unit to another. Use these as conversion factors.

1 atm =101300 Pa =101.3 kPa* =760 mmHg =

760 torr = 14.7 psi

* This is the SI unit for pressure.

Practice

Convert the following pressure units.

1.750 mm Hg to atm

2.145 kPa to mm Hg

3.1.5 atm to kPa

Temperature

The SI scale for temperature is the Kelvin scale.Kelvin (K)– temperature scale whose lowest temperature is “0”, called absolute zero.

Temperatures cannot fall below an absolute zero!!

A temperature scale with absolute zero is needed in Gas Law calculations because you can’t have negative pressures or volumes

Temperature

It is very easy to convert temperatures in oC to K through this expression:

Practice:25 oC = ___ K350 K = ___ OC0 OC = ___ K

KC 273

Standard Temperature and Pressure

Standard Temperature and Pressure (STP) = 1 atm (or the equivalent in another unit) and 0°C (or 273 K)

To ensure that scientists worldwide discuss gas behavior using the same temperature and pressure units, the international scientific community uses something called STP.

Gas Laws

The gas laws summarize observations regarding how gases behave under different conditions.

Recall that the KMT explains why gases behave as they do. The laws summarize how they behave.

“Before” and “After” in Gas Laws

4 of the gas laws we will learn have “before” and “after” conditions.

In the one of the laws we will learn, the expression is:P1V1 = P2V2

P1 and V1 represent the starting conditions (or conditions “before” they are changed) of pressure and volume for a sample of gas.

P2 and V2 represent the resulting conditions (or conditions “after” they are changed) of pressure and volume for the same gas.

Avogadro’s Law

Avogadro’s Law relates # of moles and volume when temperature and pressure are held constant.

V = Volumen = # of moles of gas

2

2

1

1

n

V

n

V

Hint: The two volume units must be the same on both sides of the equation!

Avogadro’s Law

Problem:

n1 = 0.15 mol 2.5 L = V2

V1 = 2.5 L 0. 15 mol 0.55 mol

n2 = 0.55 mol V2 = (0.55 mol)(2.5L)

V2 = ? (0.15 mol)

V2 = 9.2 L

2

2

1

1

n

V

n

V

A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles?

Practice ProblemsIf a container holds 3.6 moles of helium in 4.7 L,

how many moles could it hold if the container is increased to 5.8 L? (Hint: Use cross-multiplication to solve!)

A sample of 9.5 mol of oxygen occupies 8.2 L. What volume of the gas would be occupied by 13.0 mol?

2.8 mol of a gas are put in a container that has a volume of 15.0 L. If the size of the container is increased to 25 L, how many moles of gas could it then hold?

Boyles’ Law

Boyles’ Law relates pressure and volume when the temperature and number of moles are held constant.

P = pressureV = volume2211 VPVP The two pressure units must match and the two volume units must match!

Boyle’s Law

Problem: A 2.5 L gas sample is held at 1.05 atm. What volume will it occupy if the pressure is changed to 0.980 atm.

V1 = 2.5 L (1.05 atm)(2.5 L) = (0.980 atm)V2

P1 = 1.05 atm V2 = (1.05 atm)(2.5 L)V2 = ? (0.980 atm)P2 = 0.980 atm V2 = 2.7 L

2211 VPVP

Practice Problems

If 98 mL of a gas at 50.0 kPa has its volume reduced to 65 mL, what will the new pressure be?

A gas cylinder contains 25.0 L of gas under 5.0 atm of pressure. If the gas is transferred to a 30.0 L cylinder, what will the new pressure of the gas be?

A sample of gas has a volume of 12 L at 0.78 atm. The sample is put into a new tank under 1.4 atm of pressure. What is the volume of the tank?

Charles’ LawCharles’ Law relates temperature and Volume when pressure and number of moles are held constant.

V = VolumeT = Temperature

2

2

1

1

T

V

T

V

The two volume units must match and temperature must be in Kelvin!

Charles’ Law

Problem:

V2 = ? 10.5 L = V2

V1 = 10.5 L 298 K 323 K

T1 = 25 oC = 298 K V2 = (323K)(10.5 L)

T2 = 50 oC = 323 K 298 K

V2 = 11.4 L

2

2

1

1

T

V

T

V

What is the final volume if a 10.5 L sample of gas is heated from 25C to 50C?

Practice ProblemsA gas sample at 40 0C occupies a volume of

2.32 L. If the temperature is raised to 75 0C, what will the volume be, assuming the pressure remains constant?

A gas at 89 0C occupies a volume of 0.67 L. At what temperature will the volume be increased to 1.12 L?

What is the volume of air in a balloon that occupies 0.620 L at 25 0C if the temperature is lowered to 0 0C?

Combined Gas Law

P = PressureV = Volumen = # of molesT = Temperature22

22

11

11

Tn

VP

Tn

VP

Each “pair” of units must match and temperature must be in Kelvin!

Problem

What is the final volume if a 0.125 mole sample of gas at 1.7 atm, 1.5 L and 298 K is changed to STP when enough gas is added to increase the number of moles to 0.225 mole? V1 = 1.5 L V2 = ? n1 = 0.125 mol n2 = 0.225 mol P1 = 1.7 atm P2 = 1.0 atm T1 = 298 K T2 = 273 K

(1.7 atm)(1.5 L) = (1.0 atm)V2_ (0.125 mol)(298 K) (0.225 mol)(273 K) V2 = (0.225 mol)(273 K)(1.7 atm)(1.5 L) = 4.2 L (0.125 mol)(298 K)(1.0 atm)

22

22

11

11

Tn

VP

Tn

VP

Practice Problems

1. 3.0 mol of gas at 110 kPa and 30 0C fills a flexible container with an initial volume of 2.0 L. If the temperature is raised to 80 0C, the pressure is increased to 440 kPa, and the number of moles increases to 4.5 mol, what is the new volume?

2. At STP, a 6.5 mol gas sample occupies 30 mL. If the temperature is increased to 30 0C, the entire sample is transferred to a 20 mL container, and 3.5 additional moles of gas are added, what will be the gas pressure inside the container?

22

22

11

11

Tn

VP

Tn

VP

P1V1n1T1

P2V1n2T2

The combined gas law can be used for all of the gas law problems we have learned so far!

For example, if the temperature and number of moles are held constant, then T1 = T2 and n1 = n2. When variablesare the same, you can cancel them on both sides.

The combined gas law becomes Boyle’s Law!

The Combined Gas Law Is All You Need

Watch as variables are held constant and the combined gas law “becomes” the other 3 laws

22

22

11

11

Tn

VP

Tn

VPHold pressure and

temperature constantAvogadro’s Law

22

22

11

11

Tn

VP

Tn

VPHold moles and

temperature constantBoyles’ Law

22

22

11

11

Tn

VP

Tn

VPHold pressure and

moles constantCharles’ Law

Transforming the Combined Law

The Ideal Gas LawThe Ideal Gas Law does not compare situations—it describes a gas in one situation.

P = PressureV = Volume n = molesR = Gas Law ConstantT = Temperature

nRTPV

The value for “R” depends on the unit that is used for pressure. There are two possibilities:

Kmole

atmL

*

*0821.0

Kmole

kPaL

*

*31.8

Choose the one with units that match your pressure units!

The Ideal Gas Law Example

nRTPV Problem: A sample with 0.55 mol of gas is at 105.7 kPa and 27 oC. What volume does it occupy?

n = 0.55 mol (105.7 kPa)V = (0.55 mol)( )(300 K)P = 105.7 kPaT = 27 oC = 300 K V = (0.55)(8.31 L-kPa)(300)V = ? (105.7 kPa)R = V = 13 LKmole

kPaL

*

*31.8

Kmole

kPaL

*

*31.8

Practice Problems

1. Calculate the number of moles of gas contained in a 3.0 L vessel at 3000 K with a pressure of 1.50 atm.

2. Determine the temperature of 2.49 moles of gas contained in a 1.00 L vessel at a pressure of 143 kPa.

3. Calculate the pressure that a 0.323 mol sample of a gas have at 265 K and a volume of 7.8 L.

Kmole

atmL

*

*0821.0

Kmole

kPaL

*

*31.8R = or

What did you learn about airbags?

Airbags

States of

Matter

States of

Matter

Use different

PropertiesProperties

ChangesChanges

Gas LawsGas LawsDensityDensity

Kinetic Molecular

Theory

Kinetic Molecular

Theory

With different

Work because of changes

One of which is

GasGas

Properties explained by

To produce

Explanation for

Which is a