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Ideal Gases s Molecular Kinetic Theory 11.2.1, 11.2.2 & 11.2.3

# Ideal Gases Physics – Molecular Kinetic Theory 11.2.1, 11.2.2 & 11.2.3

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Ideal Gases

Physics – Molecular Kinetic Theory 11.2.1, 11.2.2 & 11.2.3

Gases? In our everyday lives we use atoms or molecules in the form of gases to accomplish all sorts of

jobs such as filling rubber tyres, footballs or making refrigeration systems. They have an effect on virtual everything we do on this planet and the entire universe. We even breath gases!

Mainly we call them “gases” because they have similar properties i.e. expand to fill the space they are in, are disassociated and bounce off each other and the walls of containers they are in.

Below there are two examples of gases. One is monatomic (could be a noble gas on its own), the other is diatomic (could oxygen or O2 and found in a pair). At AS /A2 level we will treat all these gases the same with respect to any formulae and rotational energies. We must only take into account the numbers of atoms if it comes up in the questions. However, if dealing in moles all would be the same for our gases formulae

NB: The idea of this is simple and based on the relative mass of carbon which is 12 a.m.u. or we define 0.012kg of carbon as consisting of 6.022 x 1023 atoms. The same goes for any element. You simply scale it up or down relative to carbon 12. It will make sense when you see an example!

Possible Physical Properties

We can examine gases by their physical attributes i.e.

Density – mass / volume ()

Pressure – force on a container wall (p)

Temperature – average kinetic energy of particles (T)

Volume – m3 (V) (1litre = 1dm3 = (10cm)3 = (0.1m)3 = 1 x 10-3m3 = 0.001m3

Mass – kg

Moles – (n) a mass & number relational system (NB)

STP – Standard Temperature / Pressure / Volume

The idea behind this is to simply create a quick reference point for any calculations. Take it to mean that;

Standard Temperature – 273.15K = 0ºC

Standard Pressure – 101.3kPa = 101300Nm-2

Volume of 1 mole of particles – 22.4 litres

When using a foot pump we find that the pump gets very hot. Reason for this is that;

Particles which have a certain kinetic energy. (directly related to Temperature)

They are squashed closer together in a smaller volume

Kinetic energy or Temperature is concentrated in a smaller area i.e. feels hotter or T rises.

Case Study?

Less Volume

Higher Temperature

NB: Consider this as an Adiabatic change. i.e. the change is so quick that no heat is lost to the surroundings

Real or Ideal?

In Science we often find that we can look at the real world or

construct a model of how things work under certain conditions. For

gases we call this “real theory” or “ideal theory”. The “ideal model

is a mathematical construct which works;

at relatively low pressures and high temperatures i.e. far from when a gas becomes a liquid. (works over a limited range)

When the “ideal model” also assumes that the intermolecular forces are very negligible and collisions are elastic (not quite true)

When the particles themselves take up no volume (obviously impossible)

Practical Investigation?When thinking about how ideal gases behave we can look towards 3 key experiments to help us establish a formulae to link together certain variables such as pressure, temperature and volume. (p, V, t)

Each experiment yielded a simple but important conclusion;

Complex Results?The most simple experiment that can be accomplished in the laboratory is the pressure law experiment;

Graph these results

Hint: What is the temperature when you extrapolate to zero pressure?

Kelvins K Pressure Nm

298 101325

303 103000

313 105000

323 109800

333 112000

343 114000

353 121700

363 122700

373 126300

Analysis – we can extrapolate back to absolute zero!

Graph to show how the Pressure of a real gas Varies with Temperature

-20000

0

20000

40000

60000

80000

100000

120000

140000

0 50 100 150 200 250 300 350 400

Temperature in Kelvins K

Pre

ssu

re in

Nm

or

Pa

Basic Results?

Results….

Using the graphical analysis method we can now say that a formulae to link together these variables must include the following;

p T or p = constant x T (Pressure law)

V T or V = constant x T (Charles Law)

p 1/V (Boyles law)

At this point we can suggest a generic formulae which fits the three as;

TconstpV

constT

pVV

Tconstp

Charles

Pressure

Boyles

NB: The idea of a constant is a fudge factor to relate this to the real world of atoms!

Results….However, if we are to find out the “constant” involved we must consider things on an atomic level;

nRTpV

Any formulae must also take into account;

1. The amount or mols present in the gas or “n”

2. An empirical number found by experimentation. Which also has the correct units to make the formulae dimensionally correct and links together differing gases across the periodic table.

R = 8.31 J K-1 mol-1

This leads us to the idea that for an ideal gas under the aforementioned conditions;

Hint: also remembered as Perv Nert!

Results….

nRTpV

• This is a tricky formulae to apply properly as the units must be applied correctly.

p = pressure in Pascal's Pa (Nm-2)

V = volume in m3

n = number of moles of gas ( 1 mol = 6.022 x 1023 atoms or molecules) NB

T = temperature in Kelvin or K (-273C = 0K)

R = Molar gas constant 8.31 J K-1 mol-1

NB: The idea if this is simple and based on the relative mass of carbon which is 12 a.m.u. or we define 0.012kg of carbon as consisting of 6.022 x 1023 atoms. The same goes for any element. You simply scale it up or down relative to carbon 12.

Example 1

V

nRTp

nRTpV

What is the pressure of a 2 moles of carbon dioxide gas at a temperature of 150K that is kept on a metal canister of volume 10 litres (10 x 10-3m3)

p = pressure in Pascal's Pa (Nm-2) ????

V = volume in m3 10 x 10-3m3

n = number of moles of gas ( 2 moles)

T = temperature in Kelvin or K 150K

R = Molar gas constant 8.31 J K-1 mol-1

p = ( 2 mols x 8.31 J K-1 mol-1 x 150K)

10 x 10-3m3

p = 249300 Jm-3

p = 249300 Nmm-3

p = 249300 Nm-2

p = 249300 Pap = 249.3 kPa

Example 2

V

nRTp

nRTpV

What is the pressure of a 3.011 x 1023 molecules of oxygen gas at a temperature of 150K that is kept in a metal canister of volume 10 litres (10 x 10-3m3)

p = pressure in Pascal's Pa (Nm-2) ????

V = volume in m3 10 x 10-3m3

n = number of moles of gas ( 1/2 mol)

T = temperature in Kelvin or K 150K

R = Molar gas constant 8.31 J K-1 mol-1

p = ( 0.5 mols x 8.31 J K-1 mol-1 x 150K)

10 x 10-3m3

p = 124650 Jm-3

p = 124650 Nmm-3

p = 124650 Nm-2

p = 124650 Pap = 124.65 kPa

Initial / Final states

A good idea to solve many problems is use the idea of initial and final states.

In the diagram below there are two states initial and final.

Each state may have the same or different values for p, V & T

Less Volume

Higher Temperature

222

111

nRTVp

and

nRTVp

Initial State Final State

Case Study?

If we also give our formulae the subscript of 1 for air in the pump and 2 for air in the ball

We can equate the two as being equal and say the following relation is correct;

This is a simplistic but useful relation and of course assumes that all the temperature is in the gas and not the pump or surroundings.

The same principle can be used when a piston in a car engine compresses air and fuel. Again assume no heat loss to the engine.

2

22

1

11

2

22

1

11

2

22

1

11

1

11 1

T

Vp

T

Vp

nT

Vp

nT

Vp

nRT

Vp

nRT

Vp

nRT

Vp

Everest Case Study…At the top of Mount Everest the temperature is around 250K, with atmospheric

pressure around 3.3 x 104 Pa. Conversely the pressure and temperature at sea level is 1.0 x 105 Pa and 300K. The density of air at sea level is found to be 1.2kgm-3. But what is the density of air at the top of Mount Everest.

1. To solve use the two situations idea and compare to compare 1 kg of gas.

2. Then draw a diagram and add the respective quantities;

Mountain - Situation - 2

Sea level Situation - 1

P1 =V1 = T1 =2 =

P2 =V2 =? T2 =2 =

Everest Case Study…This problem uses the “equating” before and after method as previously mentioned.

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