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

2. 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 O 2 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

3. 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 10 23 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 – m 3 (V) (1litre = 1dm 3 = (10cm) 3 = (0.1m) 3 = 1 x 10 -3 m 3 = 0.001m 3  Mass – kg  Moles – (n) a mass & number relational system (NB)

4. 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

5. 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

6. 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)

7. 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;

8. 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 KPressure Nm 298101325 303103000 313105000 323109800 333112000 343114000 353121700 363122700 373126300

9. Analysis – we can extrapolate back to absolute zero!

10. Basic Results?

11. 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; Charles Pressure Boyles NB: The idea of a constant is a fudge factor to relate this to the real world of atoms!

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

13. Results…. 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 m 3 n = number of moles of gas ( 1 mol = 6.022 x 10 23 atoms or molecules) NB T = temperature in Kelvin or K (-273  C = 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 10 23 atoms. The same goes for any element. You simply scale it up or down relative to carbon 12.

14. Example 1 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 -3 m 3 ) p = pressure in Pascal's Pa (Nm -2 ) ???? V = volume in m 3 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 -3 m 3 p = 249300 Jm -3 p = 249300 Nmm -3 p = 249300 Nm -2 p = 249300 Pa p = 249.3 kPa

15. Example 2 What is the pressure of a 3.011 x 10 23 molecules of oxygen gas at a temperature of 150K that is kept in a metal canister of volume 10 litres (10 x 10 -3 m 3 ) p = pressure in Pascal's Pa (Nm -2 ) ???? V = volume in m 3 10 x 10 -3 m 3 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 -3 m 3 p = 124650 Jm -3 p = 124650 Nmm -3 p = 124650 Nm -2 p = 124650 Pa p = 124.65 kPa

16. 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 Initial State Final State

17. 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.

18. Everest Case Study… At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. Conversely the pressure and temperature at sea level is 1.0 x 10 5 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 P 1 = V 1 = T 1 =  2 = P 2 = V 2 =? T 2 =  2 =

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