Gas Laws

Section 3.2

Boyle’s LawAt a constant temperature, the volume of a given mass of any gas is inversely proportional to the pressure of the gas.

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/gaslaw/boyles_law_graph_new.swf

Charles’ LawAt a constant pressure, the volume of a given mass of any gas is directly proportional to the Kelvin Temperature.

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/gaslaw/charles_law.swf

Gay – Lussac’s Law of Combining Volumes

When a gases react, the volumes consumed in the

reaction bear a simple whole number ratio to each other, and to the

volumes of any gaseous product of the reaction, if all volumes are measured

under the same conditions of temperature and

pressure.

Avogadro’s Law

Equal volumes of (ideal) gases, contain the same number of particles, or molecules, under

the same conditions.

All the Law’s Combined

Ideal Gas LawPV = nRT

Combined Gas Law1 1 2 2

1 2

P V P Vk

T T

Boyle’s LawPV k

Charles’ LawV

kT

The Combined Gas Laws

Combined Gas Law

1 1 2 2

1 2

P V P VT T

• P1, V1, and T1 are the initial pressure, volume and Kelvin temperature.• P2, V2 and T2 are the final pressure, volume and Kelvin temperature.

• Pressure can be in any units as long as it’s the same for P1 and P2.• Volume can be in any units as long as it’s the same for V1 and V2.• Temperature must be in Kelvin’s for T1 and T2.

To convert from degrees to Kelvin’s add on 273.For example 25o = 25 + 273 = 298 K

3 o

3

o

1

31

1

A sample of gas exerts a pressure of 83,326 Pa in a 300 cm vessel at 25 C.

What pressure would this gas sample exert if it were placed in a 500 cm

container at 50 C?

P 83,326 Pa

V 300cmT 25 273 298

2

32

2

1 1 2 2

1 2

2

2

2

P ?

V 500cmK T 50 273 323K

P V P VT T

83,326 300 P 500298 323

83,326 300 323P

50054,189.86 Pa P

Temperature must be in Kelvin’s

3 o

o

1

1

A sample of gas occupies 250 cm at 27 C. What volume will it occupy at

35 C if there is no change in pressure?

As the pressure is constant, it can be left out of the equation.V 300mL VT 27 273 300K

Note:

2

2

1 2

1 2

2

2

32

?T 35 273 308K

V VT T250 V300 308250 308

V300

256.67 cm P

S.T.P.(Standard, Temperature and Pressure)

• Scientists who first studied gases soon realised that the pressure and temperature controlled the volume observed for a gas sample.

• Therefore to compare different gas samples, they defined a set of reference conditions.

• These conditions are known as standard, temperature and pressure, or simply STP, and are 273 K, and 101,325 Pa.

3 o

213

1 2

1 2

1 1 1 2

1 2

1

What would the volume of a gas at STP if it was found to occupy a volume

of 255 cm at 25 C and 101,000 Pa?

P 101,000 PaP 101,325 PaV ? V 255cmT 273K T 25 273 298K

P V P VT T

101,325 V 101,000273

1

31

255298

101,000 255 273V

298 101,325V 232.86cm

Standard temperature 273 KStandardpressure 101,325 Pa

The Kinetic Theory of Gases

The kinetic theory of gases was developed by James Clerk Maxwell and Ludwig Boltzmann.

This theory assumes that:

1. Gases are made up of particles whose diameters are negligible compared to the distances between them.

2. There are no attractive or repulsive forces between these particles.3. The particles are in constant rapid random motion, colliding with

each other and with the walls of the container.4. The average kinetic energy of the particles is proportional to the

Kelvin temperature.5. All collisions are perfectly elastic .

Ideal Gases versus Real Gases

• An ideal gas is one which obeys all the gas laws and under all conditions of temperature and pressure.

• No such gases exists, but real gases behave most like an ideal gas at high temperatures and at low pressures.

• Under these conditions, the particles of a real gas are relatively far away from each other, and the assumptions of the kinetic theory are reasonably valid.

Why do real gases deviate?

• Intermolecular forces are present.(Such as dipole – dipole, Van der Waals, etc.,)

• Molecules have volume.• Collisions are not perfectly elastic.

Equation of State for an Ideal Gas

pV nRT

3

1 1

Pressure p PaVolume V m

Number of moles n molGas constant R JK molTemperature T K

Measure Symbol Unit

3 3 o

r

Volume: Temperature : no. of mols:actual mass

1L 1 10 m K C 273 nM

Conversions

o2

r

1 1

3

What volume will 24 g of O occupy at 20 C and a pressure of 89000 Pa.

p 89000PaV ?

acutal mass 24n 0.75 mols

M 32

R 8.3 J K molT 20 273 293K

pV nRT89000 V 0.75 8.3 293

0.75 8.3 293V

89000V 0.0204 m

o

o

A student collected natural gas from a laboratory gas jet at 25 C in 0.25L flask until the pressure of the

gas was 73327.30 Pa. The gas sample weighted 0.118 g at a temperature of 25 C. From this data

3 3

1 1

3

3

3

r

, calculate the molecular mass of the gas.

p 73327.30Pa

V 0.25L 0.25 10 mn ?

R 8.3 J K molT 25 273 298K

pV nRT

73327.30 0.25 10 n 8.3 298

73327.30 0.25 10n

8.3 2987.41 10 mol n

actual massM

n0

mass

3

.1187.41 10

mass 15.92 g