Balloons and Gas Laws

1. Balloons and Gas Laws Thomas L. Morton Philips Healthcare November 5, 2008

2. Introduction What will happen to our balloon? • It will rise • It will expand while rising • It will pop, and fall back to earth • We will talk about the physics of each of these steps.

3. Topics of Discussion • How do gases behave? • Why does a helium balloon rise? • What is helium? • Why does the balloon expand? • How much can it lift? • How fast does it go up? • How far does it go up?

4. How do gases behave? • Boyle’s Law – PV = const • Charles’ Law – V = const x T • Avogadro’s Law – Equal volumes at the same temperature and pressure have the same number of molecules. V = n Vstandard

7. What is the Temperature? • Volume doesn’t go to zero at 0º Celsius • Volume does go to zero at -273º Celsius • -273 is defined as Absolute Zero • We use a Temperature scale called Kelvin • º Celsius + 273 = Kelvin

8. Boyle’s Law – PV = const Charles’ Law – V = const x T Avogadro’s Law –V = n Vstandard Net result – PV / nT = R R is the gas constant R = 8.314 J / K mol R = 1.987 cal / K mol R = .08205 liter atm / K mol How do gases behave?

9. What is n? • In a typical liter of air, there are about 2.7x1022 molecules. • Scientists use a different counting scale. • A gram equivalent of a chemical is called a mole. • 1 mole = 6.02 x 1023 molecules/atoms

11. Ideal Gas Law • Finally, we have PV = nRT • P in atmospheres • V in liters • n in moles • T in kelvins • R = 0.08205 liter atm / K mol • Valid for moderate temperatures and pressures. • OK for conditions we will see

12. Why do Helium balloons float? • Why do boats float?

13. Why do Helium balloons float? • Helium balloons “float” in a sea of air. • Helium weighs less than air. • Displaces more dense air where balloon is. • Air is 78% Nitrogen, 21% Oxygen, and 1% Argon • Molecular weight of air is 0.78*28+0.21*32+.01*40 = 28.96 g/mol • “Molecular” weight of Helium is 4 g/mol

14. Let’s use the ideal gas law • Suppose I have a balloon one foot in diameter • Volume = 4πr3/3 = 4π(15.24 cm)3 / 3 = 14,800cm3 = 14.8 liters. • Weight of that air is: 14.8/(.08205*298)*28.96=17.5 gm • Weight of helium is: 14.8/(.08205*298)*4 = 2.4 gm • Lift is 17.5 –2.4 = 15.1 gm. • Rises if balloon weight is less than 15.1 gm

15. What is the density of air? • Density is mass/Volume • We can use 28.96 * n/V = 28.96*P/RT • Using P=1 atm, T=293 K (68° F), we get =28.96/(0.08205*293)=1.20 gm/liter = 1.20x10-3 gm/cc This is about 1/800 x the density of water.

16. What is the relevance to balloons? • Recall the dessicator demonstration • Tube demonstration • Measure pressure in different sized balloons • Check balloon in the freezer • Lift equation is a little different • Use Pinternal for the weight of balloon and Helium • Lift is a little less than previous slide

17. What happens as balloon rises? • Pressure drops • Temperature drops, then stays steady • Consider balloon from a couple slides ago, at 10,000 feet high • P = 700 mBar = .69 atmospheres • T = 0º C = 273 kelvins • n = .61 moles • V = nRT / P = 19.6 liters • R = 3√ 3*V/(4π) = 16.7 cm > 15.2 cm

18. What happens as balloon rises? (2) • Consider the same balloon, now at 50,000 feet • P = 120 mBar = .12 atmospheres • T = -60º C = 213 kelvins • n = .61 moles • V = nRT / P = 88.2 liters • R = 3√ 3*V/(4π) = 27.6 cm ≈ 2x15.2 cm

19. What happens as balloon rises? (3) • Consider the same balloon, now at 100,000 feet • P = 10 mBar = .01 atmospheres • T = -50º C = 223 kelvins • n = .61 moles • V = nRT / P = 1160 liters • R = 3√ 3*V/(4π) = 65.1 cm ≈ 4x15.2 cm • Balloon stretched to 18 x original surface area • Balloon thickness starts at .28 mm, goes to .015 mm

20. Sky Diode – April 2005 Flight • The Sky Diode experiment measured the brightness of the sky. As the balloon rose, the data readings initially decreased indicating that the sky got darker. However, as the balloon began to expand, the readings started to increase until the balloon popped. Once the balloon popped the readings dropped considerably. The readings then increased as the payload dropped to the ground. • In conclusion, we propose that the balloon blocked the diode’s field of vision. Therefore, the most accurate data was taken as the payload started to rise.

21. Activities • Measure balloon size, lift, and rise rate: • Use string to measure radii. • Use scale to measure weight of balloon, and lift. • Use stopwatch to measure time to rise from floor to ceiling.

22. Why do we care about rise rate? • On Oct 11, we launched two balloons • Goal was parallel flight paths • NASA balloon • 1200 gm lift • Rose at 7-800 ft/min • Surfs-up balloon • 300 gm lift • Rose at 2-300 ft/min

23. Data Analysis • What did we measure? • What did we expect? • Did it obey the trends we expected? • Can we learn anything?

24. Volume of Balloon • We measured two circumferences, call them C1 and C2. • Radii calculated by R1 = C1/2 π inches = C1 *2.54/2 π cm. • Volume can be calculated as 4 π R12 R2/3 = 4 π C12 C2 2.543 / (3*(2 π)3) cm3= C12 C2 2.543 / (6*π2) cm3 = .276726 * C12 C2 cm3

25. 12 inch balloons – Analysis • Lift is Mair-Mhelium-Mballoon • Mair =28.96*Pamb*V/R T = 28.96*.977*V/(.08205 296) • Mhelium =4.00*Pint*V/R T = 4.00*1.000*V/(.08205 · 296) • Can write above formula as: Lift=24.29*V/(.08205 · 296) - Mballoon = 1.0003*V - Mballoon

26. 12 inch balloons – Data

27. 12 inch balloons – Analysis • Terminal velocity of a balloon satisfies: v02=4*LIFT*g/ρA • LIFT is lift of balloon • g is gravitational constant 9.8 m/sec2 • ρ is density of air in kg/m3 • A is cross-sectional area of balloon • πR12

28. 12 inch balloons – Analysis • Can solve for velocity, height of balloon as a function of time: • V= v0 tanh(v0 C t) • C=ρA/4 LIFT • tanh is hyperbolic tangent • Height = ln[cosh(v0 C t) ] / C • ln is natural logarithm • cosh is hyperbolic cosine • Time to rise 5.87 meters is: arccosh[exp(5.87*C)]/ v0 C

29. 12 inch balloons – Analysis

30. 12 inch balloons – Analysis

31. 12 inch balloons – Analysis