1 / 65

13 MATTER: VERY SIMPLE The gas laws

13 MATTER: VERY SIMPLE The gas laws. Deduce the relationships between pressure, volume and temperature of a gas from experimental data Combine them to give the Ideal Gas Law. Boyle’s Law. At a constant temperature the pressure p and volume V of a gas are inversely proportional. 1627-91.

lucas
Télécharger la présentation

13 MATTER: VERY SIMPLE The gas laws

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 13 MATTER: VERY SIMPLEThe gas laws • Deduce the relationships between pressure, volume and temperature of a gas from experimental data • Combine them to give the Ideal Gas Law

  2. Boyle’s Law At a constant temperature the pressure p and volume V of a gas are inversely proportional 1627-91 pV=constant What graph could we draw to test this relationship?!?

  3. Charles’ Law V/T=constant At constant pressure, the volume V of a gas is directly proportional to its absolute temperature T. 1780’s Pressure Law p/T=constant At constant volume, the pressure p of a gas is directly proportional to its absolute temperature T.

  4. Absolute Temperature Absolute temperature measures how hot an object is on a scale starting at absolute zero (lowest possible temperature, when all particles have minimum possible energy). Measured in: Kelvin, K Scale: 1K = 1⁰C Conversion: K = C + 273 1800’s E.g.. 0 ⁰C = 273K, 10 ⁰C = -263K, -273K = 0 ⁰C

  5. Testing Charles’ Law and the pressure Law… Test the laws using the experimental setup suggested on your sheet. Take measurements of volume or pressure at 4/5 different temperatures (doesn’t matter what temps as long as good range) Plot a rough graph of your results. Don’t forget to use absolute temperature… Conversion: K = C + 273

  6. What is a mole? One mole has 6.023 x 1023 particles in it. It has a mass equal to the atomic mass in grams: 1 mole of Carbon = 12g I mole of Hydrogen = 1g

  7. One beautiful equation…. pV = nRT n is the number of moles R = molar gas constant (JK-1mol-1) Energy the gas has per Kelvin per mole

  8. Boltzmann constant is derived thus: k = R/NA JK-1per particle Energy the gas has per Kelvin per particle pV = nNAkT = nRT

  9. Summary…. pV = constant (if T is constant) V/T = constant (if p is constant) p/T = constant (if V is constant) Combine to give the Ideal Gas law… pV = nRT or pV = NkT Where… n = number of moles (amount of gas) R = Molar gas constant (8.31J mol-1K-1) k = Boltzmann constant (1.38x10-23JK-1) N = number of particles

  10. Ideal Gases Develop problem solving skills involving the gas laws. Derive the relationship density = PM/RT

  11. A sealed aluminium alloy flask contains air at atmospheric pressure and a temperature of 27 oC. The alloy’s melting point is 620 oC, and the flask will burst if the pressure exceeds 2.9 atmospheres. Will the flask melt before it bursts, or burst before it melts?

  12. Determining the density of air A plastic vessel (40 cm x 40 cm x 40 cm) containing air at atmospheric pressure is weighed. Its mass is recorded as 371.2 g. The air in the vessel is pumped out and the vessel re-sealed. Its mass is now 300.3 g. From these results, work out the density of air in kg m-3. ρ = mass/volume kgm-3 ρ = density

  13. Upthrust… Where does the force come from that causes a hot air balloon to rise? Archimedes: Any object in a fluid displaces a volume of the fluid that weighs more than the object does, will float upwards.

  14. Bye Bye Kitty……….. A sadistic physicist decides to find out how many helium party balloons need to be attached to a kitten to make it take off. He knows that the upthrust force on a helium balloon is given by the weight of air displaced by it. Mass of kitten = 0.25 kg Volume of a party balloon = 0.027 m3 Mass of balloon (excluding gas in it) = 0.005 kg Air density = 1.1 kg m-3 Helium density =0.15 kg m-3 Work out how many balloons are needed to make kitty fly! ρ = density ρ = m/V kgm-3

  15. Hint………………….. For one balloon, work out the net upward force by calculating the upthrust force and the weight of balloon plus contents. Hint: For one balloon, work out the net upward force by calculating the up thrust force and the weight of balloon plus contents. Hint: For one balloon, work out the net upward force by calculating the up thrust force and the weight of balloon plus contents.

  16. Derive… ρ = density p = pressure M = molar mass R = universal gas constant T = Absolute Temperature n = number of moles m = mass ρ = pM/RT Using the following…. pV=nRT, ρ=m/V, n=m/M

  17. Gas density Instead, you are going to derive an equation for gas density in terms of the intensive properties of pressure, molar mass and temperature, all of which are easy to determine. Write down the ideal gas equation (molar form) Write an expression for n, the number of moles in terms of the mass of gas, m, and the molar mass M. What is the standard expression for the density of a substance? Substitute equation 2 into equation 1 and rearrange to give the correct expression for density on one side. Use your new equation to calculate the density of air at atmospheric pressure (105 Pa) and 25 oC. Take the molar mass of air as 28.6 g mol-1.

  18. Kinetic theory of gases • Derive PV = nRT from a kinetic theory standpoint • Derive expressions for kinetic energy and molecular speed from the resulting equations

  19. Kinetic theory of gases

  20. Kinetic Theory A model that attempts to explain the gas laws. First some assumptions (of an ideal gas) Gas contains large number of particles Molecules move randomly at random speeds. All collisions between wall and particles are elastic. Low density so space occupied by molecules is zero Energy of motion large enough so that attractive forces can be ignored.

  21. p = mv F= Δp/Δt Momentum Recap…

  22. Impact of a ball on a wall (Qs 50S) (3mins) A ball of mass 2 kg, moving at 12 ms–1, hits a massive wall head-on and stops. 1. How much momentum did the ball have before impact? 2. How much momentum does the ball have when it has stopped, after impact? 3. How much momentum did the ball lose during impact? 4. Assume that Newton’s third law is correct and applies to this case. How much momentum did the wall gain? 24 kgms-1 0 kgms-1 24 kgms-1 24 kgms-1

  23. Force due to a stream of elastic balls Suppose the wall is hit by a stream of 2 kg balls. In this case each ball arrives with a speed of 12 m s–1and bounces straight back with an equal speed of 12 m s–1in the opposite direction. As previously, 1000 balls arrive at the wall in 10 s. 8. Calculate the change of momentum when one ball arrives at the wall and bounces away. 9. Calculate the change in momentum for all the 1000 balls. 10. Calculate the average force on the wall during that 10 s period. 48 kg m s–1 48000 kg m s–1 4.8 kN

  24. Deriving the ideal gas equation from kinetic theory Consider a cuboid, dimensions x,y,z, containing a single gas molecule of mass m travelling at speed v along the X direction. The molecule collides elastically with wall YZ and then travels in the opposite direction towards the other YZ wall. Q1. Show that the time between collisions of the molecule with the same wall is t = 2x/v. Q2. What is the rate at which the molecule collides with the wall? Q3. The molecule collides elastically with the wall and rebounds in the opposite direction. What is its change in momentum? Q4. Newton’s second law can be stated as Force = rate of change of momentum. Use this together with your answers to the previous questions to show that the average force on YZ due to the molecule colliding with it is F = mv2/x . Q5. Use the basic definition of pressure (pressure = force/area) and your answer to the last question to derive an expression for the pressure P exerted on wall YZ by the single gas molecule.

  25. Deriving the ideal gas equation from kinetic theory Now we need to extend the model to a more realistic situation where many gas molecules are colliding with the walls. Q1. If there are N molecules in the gas, how many, on average, will be travelling in the x direction? Q2. Using your expression for the pressure due to one molecule and your answer to the previous question, show that the pressure due to all molecules is given by P = Nmv2/3V . Q3. We normally write the equation from the previous question as PV = 1/3 Nmv2, where we now allow the molecules to travel at a range of speeds characterised by a mean square average speed. Can you see any similarities with the experimental ideal gas law? Q4. Write down an expression for the kinetic energy of a gas of N molecules of mass m. Q5. Combine the expression from the previous 2 questions to show that the kinetic energy of the gas is given by KE = 3/2 PV. Q6. The KE of a substance is proportional to its temperature T. Use this to show that PV α T as in the experimental ideal gas law. Q7. Show that, for molar quantities, KE = 3/2 RT = 1/2 Mv2, where M is molar mass. Q8. Use the equation from the previous question to estimate the speed of nitrogen molecules in the room around you.

  26. p = 1/3 Nmv2 / V From the 3 dimensions of the box Average of square of speeds of molecules v1 v2 Mass of one molecule v4 v3 v5 v6 v7

  27. Compare…. pV = 1/3Nm2 & pV = NkT Average KE of a molecule = 3/2kT Total KE of N molecules = 3/2NkT Kinetic energy of one mole of molecules U = 3/2RT

  28. r.m.s speed, (Root Mean Square Speed) A way of averaging speeds. The distribution of speeds of molecules in nitrogen at 300 K is shown here.

  29. r.m.s speed, Remember… Average KE of a molecule = 3/2kT So… 1/2m= 3/2kT Average of square of speeds of molecules Therefore…. = And …. vr.m.s= =

  30. (Part 2) Kinetic Theory Develop problem solving skills associated with kinetic theory of gases, including the calculation of r.m.s speeds. Starter: Estimate the speed of air molecules in this room

  31. Diffusion of gases and liquids • Observe and explain diffusion effects in liquids and gases • Derive and apply the Einstein equation for diffusion

  32. Random walk in 1 dimension Suppose a molecule can either move one step to the right (R) or one step to the left (L) as a result of collisions with neighbouring molecules. There is a 50 % chance of stepping to the right as a result of a collision, and a 50 % chance of moving to the left. Assume the distance moved in a step is always the same. Consider a molecule that starts in a certain place and then experiences 10 collisions. Q1. How many different outcomes are there for a sequence of 10 steps? (Hint: one such outcome is LRLLLRRLRL) Q2. What is the maximum distance from the start, measured in number of steps, that a molecule could end up after 10 collisions? How likely is this outcome? Q3. Consider a molecule that ends up at 8R after 10 steps. How many different ways are there of reaching this position in 10 steps? Enumerate them. Q4. What about a molecule that ends up at 6R after 10 steps? How many different ways are there of reaching this position in 10 steps? Q5. Which is the most likely place for a molecule to end up after 10 steps? Explain your answer. Q5. Can you sketch a graph to show the distribution of molecules about the starting position after 10 steps? Explain the shape of the graph you have drawn.

  33. Random walk distribution after 10 steps Gaussian or Normal distribution with standard deviation = √N = √10 = 3.2 In general, if N steps are taken, estimated distance diffused D = √N x d, where d is the “step length” or mean free path .

  34. Diffusion of perfume across a room How far do molecules of perfume diffuse in 2 seconds? Take the rms speed of the molecules as 500 ms-1, and the mean free path (step length or average distance between collisions) as 10-7 m. Hint: Work out the distance gone in 2 seconds, and then work out how many collisions are experienced in travelling this distance.

  35. A sixth form student is still recovering from a big night out and randomly walks around not knowing where he is going. Sixth form centre Estimate how many steps will it take for a student to randomly walk from common room to this lab Lab

  36. Diffusion of photons through the radiative zone of the Sun Photons that are created during fusion in the core of the Sun can take a very long time to reach the surface and be emitted. This is because they undergo a vast number of collisions as they are scattered by protons, electrons and other particles in the radiative zone. In this exercise, your task is to estimate how long it takes a photon to traverse the radiative zone as it undergoes many randomising collisions.

  37. Diffusion in the Sun Data: Depth of radiative zone: 109 m Mean free path of photon: 0.01 m (distance between successive collisions) Speed of photon: 3 x 108 ms-1 1 year = 3 x 107 seconds Q1. Using the random walk model, how long does it take a photon to travel through the radiative zone? Q2. Will more photons take a greater or less time to transit the radiative zone than the value calculated in Q1? Q3. What factors will affect the mean free path of a photon, and how will they affect it? Q4. Predict and explain the effect of increasing the mean free path on the time taken to transit the radiative zone.

  38. Internal energy and specific thermal capacity • Explain and use the equation E = mc • Explain the consequences of the anomalous STC of water • Explain the pattern in molar STC values for metals • Apply the First Law to thermodynamic problems Starter: What is the kinetic energy of 1 mole of an ideal gas at 300 K? Use R = 8.3 J K-1 mol-1. How much energy would you need to supply to raise the temperature of 1 mole of an ideal gas by 1 degree K?

  39. Compare…. pV = 1/3Nm2 & pV = NkT Average KE of a molecule = 3/2kT Total KE of N molecules = 3/2NkT Kinetic energy of one mole of molecules U = 3/2RT

  40. Molar Specific Thermal Capacity U = 3/2 RT (internal energy) Molar STC = dU/dT= 3/2 R “The molar specific thermal capacity (C) of a substance is the amount of energy needed to raise the temperature of 1 mol of substance by 1K (1⁰C)” Unit: J mol-1 K-1

  41. Specific Thermal Capacity “The specific thermal capacity (c) of a substance is the amount of energy needed to raise the temperature of 1kg of substance by 1K (1⁰C)” Unit: Jkg-1K-1

  42. How much energy to have a bath? What do we need to know? • Mass of water • STC of water • Temp change ΔE=mcΔθ (at constant volume only)

More Related