Gas

Gas

A gas is made up of moving atoms and molecules.

The pressure exerted by the gas is related to the collisions of the its molecules with the wall of its container.

The ability of a gas to fill the volume of its container must surely be due to the freedom of motion of its molecules.

The temperature and internal energy of the gas is related to the kinetic energy of these molecules.

Avogadro’s number

1 mole is the number of atoms in a 12 g sample of carbon-12.

The number of atoms in 1 mole is NA= 6.02 1023

where NA is called Avogadro’s number.

The number of moles,

where N is the number of molecules.

The number of moles,

where M is the molecular mass of the substance (mass of 1 mole) and msam is the mass of the substance in g.

AN

Nn

M

mn sam

Molecular masses of some common gases

Substance Symbol Molecular Mass (g/mol)

Molecular Hydrogen

H2 2

Helium He 4

Water vapour H2O 18

Neon Ne 20

Molecular Nitrogen N2 28

Molecular Oxygen O2 32

Argon Ar 40

Carbon dioxide CO2 44

Mole A mole is the amount of material whose mass,

m, in grams is numerically equal to the molecular mass, M, of the substance.

For example 1 mole of oxygen gas is 32 g. 1 mole of oxygen gas contain 6.02 1023 of

oxygen gas molecules.

1 mole of gas at 273.15 K occupies a volume of 22.4 L at pressure 101 kPa. (STP)

Boyle’s Law

For a fixed mass of gas at constant temperature, the product of pressure and volume is constant.

PV = constant

At constant temperature,Pi Vi = Pf Vf

P

V

Example

A cylinder with height of 0.20 m and a cross-sectional area of 0.040 m2 has a close-fitting piston that may be moved to change the initial volume of the cylinder. Air at atmospheric pressure ( 1.01 x 105 N m-2) fills the cylinder. If the piston is pushed until it is within 0.12 m of the end of the cylinder, what is the new pressure of the air. Assume that the temperature of the gas remains constant.

Solution

From Boyle’s Law:P1V1 = P2V2

Pa

V

VPP

5

5

2

112

107.1

)040.012.0(

)040.020.0)(1001.1(

Charles’ Law For a fixed mass of gas at constant pressure, the

volume is directly proportional to the absolute temperature.

constant

At constant pressure

T

V

f

f

i

i

T

V

T

V

Graphs of Charles’ Law

-273 0 T(oC)

Volume Volume

0 T(K)

The Pressure Law

For a fixed mass of gas at constant volume, the pressure is directly proportional to the temperature measured in Kelvins.

= constant.

, constant volumef

f

i

i

T

P

T

P

T

P

Example

An automobile tire is filled to a gauge pressure of 240 kPa in the morning when the temperature is 15 0C. After the car is driven all day over hot road, the tire temperature is 70 0C. Estimate the new gauge pressure.

Solution

New gauge pressure is 305 kPa.

kPa

K

KkPakPa

T

TPP

T

P

T

P

406

)15273(

)70273()101240(

1

212

2

2

1

1

Ideal Gas Law

Ideal gas Law: PV = nRT

where R (= 8.31 J mol-1 K-1) is called the gas constant

Ideal gas Law: PV = NkT

where k (= 1.38x10-23 J k-1) is called Boltzmann constant

k = R/NA

Note temperature expressed in Kelvin.

Example

A spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having volume of 125.00 cm3 is at 22 0C. It is then tossed into an open fire. When the temperature of the gas in the can reaches 195 0C, what is the pressure inside the can? Assume any change in the volume of the can is negligible.

Solution

Since no gas escape and the volume is the same.

Ideal gas law can be expressed as

kPa

PT

TP

T

P

T

P

ii

ff

f

f

i

i

320)202)(295

468(

)(

Example

An ideal gas occupies a volume of 100 cm3 at 20 0C and 100 Pa. Find the number of mole of gas in the container.

Solution

Volume, V = 100 cm3 = 1.00 x 10-4 m3. Temperature, T = 20 0C = 293 K. The number of moles,

mol

RT

PVn

6

4

1011.4

)293)(314.8(

)1000.1)(100(

Example

A cylinder contains 12 L of oxygen at 20 0C and 15 atm. The temperature is raised to 35 0C, and the volume is reduced to 8.5 L. What is the final pressure of the gas in atmosphere? Assume the gas is ideal.

Solution

Ti = 273 + 20 = 293 K

Tf = 273 + 35 = 308 K

atmVT

TVPP

T

VP

T

VP

fi

fiif

f

ff

i

ii

22)5.8)(293(

)308)(12)(15(

Example

What is the density of carbon dioxide gas at a temperature of 23 oC and atmospheric pressure?

Solution

Mass, m, is related to the number of moles by

m = nM. For carbon dioxide, M = 44 g mol-1

Density

3-m kg

RT

MP

PnRT

nM

V

m

81.1)23273)(31.8(

)1001.1)(1044(

/

53

FEQ Sept 05 (Q3(b))

At 25.0 m below the surface of the sea (sea = 1025

kg/m3), where the temperature is 5.00 oC, a diver exhales an air bubble having a volume of 1.00 cm3. The surface temperature and pressure are 20.0 oC and 1.01 × 105 Pa, respectively.

(i) State the ideal gas law and describe all the notations used. [2 marks]

(ii) Determine the volume of the air bubble just before it breaks the surface. [6 marks]

Solution Q3(bii)Determine at depth 25.0m,

(1) the absolute pressure, P1 = Patm + seagh = Pa

(2) the absolute temperature, T1 = K

State volume in SI units, V1 = m3

Surface temperature, T2 = K and

Pressure, P2 = 1.01 × 105 Pa

Assume constant number of molecules, n

21

211

2

22

1

11

PT

TVPV

T

VP

T

VP

f

Dalton’s Law of Partial Pressure

The total pressure of a mixture of gases, which do not interact chemically , is equal to the sum of the partial pressure, i.e. to the sum of the pressures that each gas would exert if it alone occupied the volume containing the mixture.

Suppose a volume contains n1 moles of gas at partial pressure P1 and n2 moles of gas at partial pressure P2.

Total pressure P = P1 + P2

Example

Given 2 moles of H2, 4 moles of O2 and 6 moles of He in a 5-liter vessel at 27 oC, determine the partial pressure of each gas and the total pressure of the mixture.

Solution Temperature, T = 27 + 273.15 = 300.15 K Partial pressure can be determine from the ideal gas law,

Total pressure, Ptotal = 59.3 105 Nm-2

V

nRTP

253

1098.9105

15.300315.822

NmPH

253

108.19105

15.300315.842

NmPO

253

106.29105

15.300315.86

NmPHe

Example

Determine the partial pressure of each gas in a mixture made up of 6g of H2, 32g of O2 and 56g of N2 if the barometric pressure is 750 torr?

Solution

The number of moles of

H2: n1 = (6/2) = 3 mol

O2: n2 = (32/32) = 1 mol

N2: n3 = (56/28) = 2 mol

Total number of moles, n = 6 mol

Partial pressure for

H2: P1 = (3/6)(750 torr) = 375 torr

O2: P2 = 125 torr

N2: P3 = 250 torr