Section 3.7—Gas Laws

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

Pressure Units

Several units are used when describing pressure

Unit Symbol

atmospheres atm

Pascals, kiloPascals

millimeters of mercury

pounds per square inch

Pa, kPa

mm Hg

psi

1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi

Definition

Kelvin (K)– temperature scale with an 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

KC 273

Definition

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

Problems often use “STP” to indicate quantities…don’t forget this “hidden” information when making your list!

Gas Laws

KMT and Gas Laws

The Gas Laws are the experimental observations of the gas behavior that the Kinetic Molecular Theory explains.

“Before” and “After” in Gas Laws

This section has 4 gas laws which have “before” and “after” conditions.

For example:

2

2

1

1

n

P

n

P

Where P1 and n1 are pressure and # of moles “before”

and P2 and n2 are pressure and # of moles “after”

Both sides of the equation are talking about the same sample of gas—with the “1” variables before a change, and the “2” variables after the change

Avogadro’s Law

Avogadro’s Law relates # of particles (moles) and volume.

Where Temperature and Pressure are held constant

V = Volumen = # of moles of gas

2

2

1

1

n

V

n

V

Example: 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?

The two volume units must match!

Avogadro’s Law

Avogadro’s Law relates # of particles (moles) and volume.

Where Temperature and Pressure are held constant

V = Volumen = # of moles of gas

2

2

1

1

n

V

n

V

Example: 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?

The two volume units must match!

n1 = 0.15 moles

V1 = 2.5 L

n2 = 0.55 moles

V2 = ? L

mole

V

mole

L

55.015.0

5.2 2

215.0

5.255.0V

mole

Lmole

V2 = 9.2 L

Boyles’ Law

Boyles’ Law relates pressure and volumeWhere temperature and # of molecules are held constant

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

Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 0.980 atm?

Boyles’ Law

Boyles’ Law relates pressure and volumeWhere temperature and # of molecules are held constant

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

Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 0.980 atm?

P1 = 1.05 atm

V1 = 2.5 L

P2 = 0.980 atm

V2 = ? LV2 = 2.7 L

2980.05.205.1 VatmLatm

2980.0

5.205.1V

atm

Latm

Charles’ LawCharles’ Law relates temperature and pressure

Where pressure and # of molecules 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!

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

V1 = 10.5 L

T1 = 25C

V2 = ? L

T2 = 50C

Temperature needs to be in Kelvin!

25C + 273 = 298 K

50C + 273 = 323 K

Charles’ LawCharles’ Law relates temperature and pressure

Where pressure and # of molecules 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!

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

V1 = 10.5 L

T1 = 25C

V2 = ? L

T2 = 50C V2 = 11.4 L

= 298 K

= 323 K

K

V

K

L

323298

5.10 2 2298

5.10323V

K

LK

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!

Example: 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 and particles are added to 0.225 mole?

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!

Example: 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 and particles are added to 0.225 mole?

P1 = 1.7 atm

V1 = 1.5 L

n1 = 0.125 mole

T1 = 298 K

P2 = 1.0 atm

V2 = ? L

n2 = 0.225 mole

T2 = 273 K

V2 = 4.2 L

STP is standard temperature (273 K) and pressure (1 atm)

Kmole

Vatm

Kmole

Latm

273225.0

0.1

298125.0

5.17.1 2

2298125.00.1

5.17.1273225.0V

Kmoleatm

LatmKmole

22

22

11

11

Tn

VP

Tn

VP

22

12

11

11

Tn

VP

Tn

VP

The combined gas law can be used for all “before” and “after” gas law problems!

For example, if volume is held constant, then

and the combined gas law becomes:

When two variables on opposites sides are the same, they cancel out and the rest of the equation can be used.

21 VV

22

2

11

1

Tn

P

Tn

P

Why you really only need 1 of these

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 Law

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

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

nRTPV

There are two possibilities for “R”:

Kmole

atmL

*

*0821.0

Kmole

kPaL

*

*31.8

Choose the one with units that match your pressure units!

Volume must be in Liters when using “R” to allow the unit to cancel!

The Ideal Gas Law Example

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

P = PressureV = Volume (in L)n = molesR = Gas Law ConstantT = Temperature

nRTPV

Example: A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy?

The Ideal Gas Law Example

Example: A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy?

n = 0.55 moles

P = 105.7 kPa

T = 27°C + 273 = 300 K

V = ?

R = 8.31 L kPa / mole K

)300(**31.8)55.0()7.105( KkmolekPaLmoleVkPa

V2 = 13 L

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

P = PressureV = Volume (in L)n = molesR = Gas Law ConstantT = Temperature

)7.105(

)300(**31.8)55.0(

kPa

KkmolekPaLmole

V

nRTPV

Chosen to match the kPa in the “P” above

Example: What is the final volume if a 15.5 L sample of gas at 755 mm Hg and 298 K is changed to STP?

Let’s Practice

Example: What is the final volume if a 15.5 L sample of gas at 755 mm Hg and 298 K is changed to STP?

P1 = 755 mm Hg

V1 = 15.5 L

T1 = 298 K

P2 = 760 mm Hg

V2 = ? L

T2 = 273 K

V2 = 14.1 L

“moles” is not mentioned in the problem—therefore it is being held constant.It is not needed in the combined law formula.

K

VHgmm

K

LHgmm

273

760

298

5.15755 2

2298760

5.15755273V

KHgmm

LHgmmK

STP is standard temperature (273 K) and pressure (1 atm or 760 mm Hg)

22

22

11

11

Tn

VP

Tn

VP

Let’s Practice

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