Number of lone electron pairs0 1 2 3

3

trigonal planar bent    4

tetrahedraltrigonal

pyramidal bent  5

trigonal bipyramidal see-saw T-shaped linear

6

octahedralsquare

pyramidal square planar T-shaped

Tota

l n

um

ber

of

ele

ctro

n p

air

sRecap VSEPR Theory

2

Gases• Gases assume the volume and shape of their

containers.

• Gases are the most compressible state of matter.

• Gases will mix evenly and completely when confined to the same container.

• Gases have much lower densities than liquids and solids

• Boyle’s Law - the volume of a gas is inversely proportional to the pressure it exerts:

V = kB (1/P) or PV = kB

Empirical Gas Laws

• Charles’ Law - the volume of a gas is directly proportional to its absolute temperature

V = kCT

• Avogadro’s Law – equal volumes of gas at the same pressure and temperature contain equal number of molecules

V = kA n 3

• Charles’ Law predicts that the volume of a gas keeps decreasing as the temperature is lowered

• The temperature at which the V is zero and the gas occupies no space is called absolute zero: it is the lowest possible temperature

Charles’ Law and Absolute TemperatureV = k2T

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• On the Kelvin scale, this temperature is set as O K:

• 0 K is the same as -273 oC

• 100. K is the same as (-273 + 100.) = -173 oC

• 0 oC is the same as 273 K

Ideal Gas Equation

pressure

volume number ofmoles

absolutetemperature

gasconstant

PV = nRT

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PV = kB

V = kCTV = kA n

Ideal Gas Equation

PV = nRT if n and T are fixed then

(Boyle’s Law)

PV = nRTif n and P are fixed then

(Charles’ Law)

PV = nRTif T and P are fixed then

V = nRT / P

V = (nR/P ) × T

(Avogadro’s Law)

V = (RT/P ) × n

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unit atmospheric pressure

pascal (Pa)kilopascal (kPa)

1.01325105 Pa 101.325 kPa

atmosphere 1.00 atm

mm mercury = torr 760 torr

pounds/sq inch (psi) 14.7 lb in–2

bar 1.01325 bar

• Most convenient unit is the atmosphere (atm)

Common Units of Pressure

• SI unit is the Pascal (Pa)

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R = 8.314 J mol–1 K–1 or Pa m3 mol–1 K–1

The Gas Constant R

Use if you are using SI units e.g. pressure in Pa

R = 0.08206 atm L mol–1 K–1

Use if you are using atmospheres and litres

PV = nRT

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Q: The lung capacity of an average human born & living at sea level is ~6 L. How many moles of air will be present in full lungs at 25 C and 1.00 atm?

A: Volume and pressure has been provided in

L and atm so use R = 0.08206 atm L mol–1 K–1

Example: Lung Capacity

RT

PVn = =

(0.08206 atm L mol–1 K–1 x 298 K)

(1.00 atm) x (6 L)= 0.3 mol

PV = nRT

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Effect of Changing VolumePV = nRT

• If the size of a sealed container is changed then n, R and T do not change

• If the initial pressure and volume are Pi and Vi and the new pressure and volume are Pf and Vf

PiVi = PfVf

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Gas Mixtures: Partial Pressures

• Dalton’s Law: the total pressure (PT) is the sum of the pressures due to the individual gases, Pi:

PT = PA + PB + PC + ….

• Air is a mixture of gases with ~79% N2 and ~21% O2:

• At atmospheric pressure, PT = 1.0 atm so:PN2

= 0.79 x 1.0 atm = 0.79 atmPO2

= 0.21 x 1.0 atm = 0.21 atm

• At a depth of 50 m, PT = 6.0 atm so:PN2

= 0.79 x 6.0 atm = 4.7 atmPO2

= 0.21 x 6.0 atm = 1.3 atm

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Gas Stoichiometry

• V is proportional to the number of moles:2 mol of H2(g) requires 1 mol of O2(g)2 volumes H2(g) requires 1 volumes of O2(g)

2H2(g) + O2(g) 2H2O(l)

V = (RT/P ) × n

• If we react 2 L of H2 with 1 L of O2, then all of the reactants will react

• If we react 1 L of H2 with 1 L of O2, then H2 is the limiting reagent and O2 will be left over at the end of the reaction

• If we react 2 L of H2 with 0.5 L of O2, then O2 is the limiting reagent and H2 will be left over at the end of the reaction

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Gas Stoichiometry

Q: What is the volume of CO2 produced at body temperature (37 oC) and 1.00 atm when 5.60 g of glucose is burnt in air?

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

A: (i) Convert mass into moles:number of moles = m / M = 5.60 g / 180.2 g mol-1

= 0.0311 mol of glucose

(ii) Use chemical equation for stoichometry:number of moles of CO2(g) = 6 x 0.0311 = 0.187

mol

(iii) Use ideal gas law to calculate volume:V = nRT / P = (0.187)(0.082606)(37 + 273)/(1.00)

= 4.76 Ldon’t forget to convert to

Kelvin!

• By the end of this lecture, you should:− be able to describe how the volume of a gas

varies with pressure, temperature and the number of molecules present

− be able to use the ideal gas equation and choose the appropriate value of R to use

− be able to work out and use partial pressures for gas mixtures

− be able to complete the worksheet (if you haven’t already done so…)

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Learning Outcomes:

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Questions to complete for next lecture:

1. One mole of gas occupies 22.414 L at STP (1 atm and 0 oC). What is the volume in m3 occupied by one mole of gas at STP?

(Hint : 1 m3 = 1000 L)

2. Use the ideal gas law and your answer to Q1 to work out the value of the gas constant, R, and its units when volume and pressure are given in S.I. units (m3 and Pa respectively).(Hint: from the data sheet, 1 atm = 1.013 105 Pa.) 

  

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Questions to complete for next lecture:

3. A 2.0 L pressure cooker is sealed at atmospheric pressure and 25 oC. What will the pressure in the cooker be at 120 oC?

4. A 12 L air cylinder is filled to a pressure of 200. atm in an air conditioned diving shop at 22 °C. What will be the pressure inside the tank once it has been left in the sun at 35 °C?