1. Thermal Physics IB Physics Topic 3: Ideal gases

2. Ideal Gases Understand and Apply the following. Understand and Apply the following. Pressure. Pressure. Gas Laws (by name) Gas Laws (by name) PV = nRT PV = nRT Kinetic Molecular Theory Kinetic Molecular Theory Explain Pressure Explain Pressure WilliamThompson (Lord Kelvin)

3. Pressure Pressure is defined as force per unit area Pressure is defined as force per unit area Newtons per square metre or N/m 2 Newtons per square metre or N/m 2 1 Nm -2 = 1 Pa (pascal) 1 Nm -2 = 1 Pa (pascal) The weight of the person is the force applied to the bed and the small area of each nail tip combines to make an overall large area. The weight of the person is the force applied to the bed and the small area of each nail tip combines to make an overall large area. The result is a small enough pressure which does not puncture the person. The result is a small enough pressure which does not puncture the person. Click on Me

4. Atmospheric Pressure Basically weight of atmosphere! Basically weight of atmosphere! Air molecules are colliding with you right now! Air molecules are colliding with you right now! Pressure = 1.013 x10 5 N/m 2 = 14.7 lbs/in 2 ! Pressure = 1.013 x10 5 N/m 2 = 14.7 lbs/in 2 ! Example: Sphere w/ r = 0.1 m Example: Sphere w/ r = 0.1 m Demo Demo A = 4  r 2 =.125 m 2 F = 12,000 Newtons (over 2,500 lbs)! 21

5. Qualitative Demonstration of Pressure y 16 Force due to molecules of fluid colliding with container. Force due to molecules of fluid colliding with container. Force α Impulse =  p Force α Impulse =  p Average Pressure = F / A Average Pressure = F / A

6. Pressure Pressure is defined as force per unit area Pressure is defined as force per unit area Newtons per square metre N/m 2 Newtons per square metre N/m 2 The pressure exerted by a gas results from the atoms/ molecules “bumping” into the container walls The pressure exerted by a gas results from the atoms/ molecules “bumping” into the container walls More atoms gives more bumps per second and higher pressure More atoms gives more bumps per second and higher pressure Higher temperature means faster atoms and gives more bumps per second and higher pressure Higher temperature means faster atoms and gives more bumps per second and higher pressure At sea level and 20°C, normal atmospheric pressure is At sea level and 20°C, normal atmospheric pressure is 1atm ≈ 1 x 10 5 N/m 2 1atm ≈ 1 x 10 5 N/m 2

7. Gases Gases (as we will see) can behave near perfectly. Gases (as we will see) can behave near perfectly. N A = 6.02 x 10 23 molecules mol -1 N A = 6.02 x 10 23 molecules mol -1 Molecules size ~ 10 -8 m to 10 -10 m Molecules size ~ 10 -8 m to 10 -10 m Example: How molecules are there in 6 grams of hydrogen gas? We have 3 moles, H 2 has 2 grams mol -1 We have 3 moles, H 2 has 2 grams mol -1 3 x 6.02 x 10 23 = 1.81 x 10 24 molecules. 3 x 6.02 x 10 23 = 1.81 x 10 24 molecules.

8. Example Make a rough estimate of the number of water molecules in an ordinary glass of water. A glass contains about 0.3 L of water, which has a mass of about 300 g. Since the molar mass of water (H 2 O) is 18 g mol -1 A glass contains about 0.3 L of water, which has a mass of about 300 g. Since the molar mass of water (H 2 O) is 18 g mol -1 Hence, 300g/18g mol -1 ~ 17 mol ~ 10 25 molecules Hence, 300g/18g mol -1 ~ 17 mol ~ 10 25 molecules

9. The Boyle-Mariotte Law Gases (at constant temperature) decrease in volume with increasing pressure. Gases (at constant temperature) decrease in volume with increasing pressure. P =F/A P =F/A V = πr 2 h V = πr 2 h

10. Figure 17-3 Increasing Pressure by Decreasing Volume

11. The Boyle-Mariotte Law Gases (at constant temperature) decrease in volume with increasing pressure. Gases (at constant temperature) decrease in volume with increasing pressure. Isothermal Process Isothermal Process PV = constant PV = constant P 1 V 1 = P 2 V 2 P 1 V 1 = P 2 V 2

12. Example The pressure of gas is 2 atm and its volume 0.9 L if the pressure is increased to 6 atm at constant temperature, what is the new volume? The pressure of gas is 2 atm and its volume 0.9 L if the pressure is increased to 6 atm at constant temperature, what is the new volume? Answer: Answer: P 1 V 1 = P 2 V 2 2 x 0.9 = 6 x V; hence V = 0.3 L

13. The volume-temperature law Charles & Gay-Lussac Charles & Gay-Lussac Isobaric Process Isobaric Process V/T = constant V/T = constant V 1 / T 1 = V 2 / T 2 V 1 / T 1 = V 2 / T 2 At absolute zero a gas would take up zero volume. In reality they liquefy when they get really cold! At absolute zero a gas would take up zero volume. In reality they liquefy when they get really cold!

14. The pressure-temperature Law Gases (at constant volume) increase in temperature with increasing pressure. Gases (at constant volume) increase in temperature with increasing pressure. Isochoric Process Isochoric Process P/T= constant P/T= constant P 1 /T 1 = P 2 /T 2 P 1 /T 1 = P 2 /T 2 0100 200-200-100 temp. °C pressure

15. Example A bottle of hair spray is filled to a pressure of 1 atm at 20°C A bottle of hair spray is filled to a pressure of 1 atm at 20°C What is the canister pressure if it is placed into boiling water, and allowed to reach thermal equilibrium? What is the canister pressure if it is placed into boiling water, and allowed to reach thermal equilibrium? P 1 / T 1 = P 2 / T 2 or p 1 T 2 = p 2 T 1 1 / 293 = p 2 / 373 p 2 = 373/293 p 2 = 1.27 atm

16. Absolute zero Ideal gas has zero volume Ideal gas has zero volume Resistance of metal drops to zero (actually superconductivity cuts in above 0K) Resistance of metal drops to zero (actually superconductivity cuts in above 0K) Brownian motion ceases! (kinetic energy = 3/2 kT) Brownian motion ceases! (kinetic energy = 3/2 kT) But lowest temperature attained is ≈ 10 -9 K But lowest temperature attained is ≈ 10 -9 K 0 -273 °C temp. °C pressure

17. Equations of state State, identifies whether solid liquid or gas State, identifies whether solid liquid or gas Key parameters or state variables Key parameters or state variables Volume, V (m 3 ) Volume, V (m 3 ) Pressure, p (N/m 2 ) Pressure, p (N/m 2 ) Temperature, T (K) Temperature, T (K) Mass, M (kg) or number of moles, n Mass, M (kg) or number of moles, n Equation of state relates V, p, T, m or n Equation of state relates V, p, T, m or n

18. Bulk vs molecules Consider force between two molecules Consider force between two molecules At absolute zero At absolute zero No thermal energy No thermal energy Molecules sit at r 0 Molecules sit at r 0 Above absolute zero Above absolute zero Some thermal energy Some thermal energy Molecules are at r> r 0 (thermal expansion) Molecules are at r> r 0 (thermal expansion) At high temperature At high temperature Thermal energy > binding energy Thermal energy > binding energy Molecules form a gas Molecules form a gas r force energy r0r0 repulsion attraction binding energy thermal energy

19. Equation of state for a gas All gases behave nearly the same All gases behave nearly the same pV = nRT pV = nRT R = 8.3 J/(mol K) for all gases (as long as they remain a gas) R = 8.3 J/(mol K) for all gases (as long as they remain a gas) T is in K!!!!!! T is in K!!!!!!

20. Example What is the mass of a cubic metre of air? What is the mass of a cubic metre of air? Molecular weigh of air ≈ 32g Molecular weigh of air ≈ 32g pV = nRT Atmospheric pressure = 10 5 N/m 2 Atmospheric temp. = 300K For a volume of 1 m 3 n = pV/RT = 10 5 / (8.3 x 300) = 40 moles M = 40 x 0.032 = 1.3kg

21. Constant mass of gas For a fixed amount of gas, its mass or number of moles remains the same For a fixed amount of gas, its mass or number of moles remains the same pV/T = nR = constant pV/T = nR = constant Comparing the same gas under different conditions Comparing the same gas under different conditions p 1 V 1 /T 1 = p 2 V 2 /T 2 p 1 V 1 /T 1 = p 2 V 2 /T 2 Hence can use pressure of a constant volume of gas to define temperature (works even if gas is impure - since all gases the same) Hence can use pressure of a constant volume of gas to define temperature (works even if gas is impure - since all gases the same) Must use T in K!!!!!! Must use T in K!!!!!!

22. Example A hot air balloon has a volume of 150m 3 A hot air balloon has a volume of 150m 3 If heated from 20°C to 60°C how much lighter does it get? If heated from 20°C to 60°C how much lighter does it get? Molecular weight of air ≈32g Molecular weight of air ≈32g pV/T = nR n = pV/RT Balloon has constant volume and constant pressure n cool =10 5 x150 / (8.3 x293) = 61680 n hot =10 5 x150 / (8.3 x333) = 54271  n = 7409 moles  M = 7409 x 0.032 = 237kg

23. Molecules have finite size Cannot reduce volume of gas to zero! Cannot reduce volume of gas to zero! When you try, it becomes a liquid When you try, it becomes a liquid Slightly increases the measured volume Slightly increases the measured volume Atoms/ molecules always attract each other Atoms/ molecules always attract each other Slightly reduces the measured pressure Slightly reduces the measured pressure Van de Waals forces Van de Waals forces

24. p-V diagrams (for gases) Useful to consider the pressure/volume changes at constant temperature Useful to consider the pressure/volume changes at constant temperature Isotherms are p-V values for a fixed amount of gas at constant volume Isotherms are p-V values for a fixed amount of gas at constant volume p  1/V p  1/V volume Pressure Increasing temperature

25. Kinetic theory of gases A gas consists of a large number of molecules. A gas consists of a large number of molecules. Molecules move randomly with a range of speeds. (Maxwell's Distribution) Molecules move randomly with a range of speeds. (Maxwell's Distribution) The Volume of the molecule is negligible compared with the volume of the gas itself. The Volume of the molecule is negligible compared with the volume of the gas itself. Collisions are elastic (KE conserved) Collisions are elastic (KE conserved) No inter-molecular forces. No inter-molecular forces. Collision time with walls are very smal. Collision time with walls are very smal. Molecules obey Newton’s Laws of Mechanics Molecules obey Newton’s Laws of Mechanics

26. Molecules in a gas Gas atoms/molecules move in a straight line Gas atoms/molecules move in a straight line velocity due to thermal energy velocity due to thermal energy KE = 1/2 m v x 2 ≈ 3/2 kT KE = 1/2 m v x 2 ≈ 3/2 kT KE avg absolute temp KE avg α absolute temp RMS – (Root mean squared) RMS – (Root mean squared) Pressure results from collisions with the walls of the container. (NOT collisions between molecules Pressure results from collisions with the walls of the container. (NOT collisions between molecules F impact = Δp/t = (2m v x )/t F impact = Δp/t = (2m v x )/t

27. Example How fast does a typical gas molecule (travel at room temperature? Lets take Nitrogen-14! (k = 1.38x10 -23 J/K) How fast does a typical gas molecule (travel at room temperature? Lets take Nitrogen-14! (k = 1.38x10 -23 J/K) KE = 1/2 mv 2 = 3/2 kT v = (3kT/m) 1/2 v = [(3)(1.38x10 -23 x 293/(4.65x10 -26 )] 1/2 v = 511 m/sec

28. Example If it takes 2 mins for your kettle to begin boiling how much longer does it take to boil dry? If it takes 2 mins for your kettle to begin boiling how much longer does it take to boil dry? Assume kettle is 3kW Assume kettle is 3kW Starting temp of water 20°C Starting temp of water 20°C Work done by kettle = power x time = 2 x 60 x 3000 = 360 000J = Work to boil water of mass M =  T x M x c water 360 000J = 80 x M x 4190 Mass of water = 1.07kg Energy to boil water = M x L v (water) = 1.07 x 2256 x10 3 = 2420 000J Time required = Energy /power = 2420 000/3000 = 808 s ≈ 13mins