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The Ideal Gas

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The Ideal Gas LectureLecture 2222 Equations of state. The ideal gas. Kinetic model of a gas. Phase Diagram State variables State variables for a system: they characterize the system macroscopically at a given time. • Pressure • Temperature • Volume • Mass or moles Important: a system is in a thermal state that can be described by these variables when ALL the system has uniform pressure, temperature, etc. Example: Gas in a tube whose ends are kept at different temperature is NOT in a thermal state. Equation of state The equation of state relates the state variables. Can be obtained: • Empirically • From statistical mechanics The ideal gas An ideal gas is one whose molecules • are very small (point-like) • do not interact with one another This is very true for any gas that is not in an “extreme” situation. The gas is NOT ideal if we have: • high density • high pressure • near transition to liquid (or solid) • molecules are very large Mole of a gass A mole (mol) is defined as the number of grams of a substance that is numerically equal to the molecular mass of the substance: 1 mol of H2 has a mass of 2 g (2*1 protons + 2*0 neutrons per moleule) 1 mol of Ne has a mass of 20 g (1*10 protons + 1*10 neutrons per atom) 1 mol of O2 has a mass of 32 g (2*8 protons + 2*8 neutrons per molecule) 1 mol of C has a mass of 12 g (1*6 proton + 1*6 neutrons per atom) 1 mol of CO2 has a mass of 44 (1*6+2*8) protons + (1*6+2*8) neutrons per molecule mass Number of moles: n = molecular mass Q: How do I know the molecular mass? A: Look at the periodic table. The ideal gas state equation N N = = 6.02×1023 Avogadro's number pV = NkT A n J R = 8.314 Ideal gas constant OR mol K R − pV =nRT k = = 1.38×10 23 J/K Boltzmann's constant NA N = number of molecules/atoms 23 n = number of moles 1 mole of a atoms = 6.022х10 atom Avogadro’s number: 6.022х1023 Constant p Constant T ∝ V T (In kelvins) 1 p ∝ V Example: Volume of ideal gas at STP STP = Standard Temperature and Pressure = 0°C and 1 atm What is the volume of one mole of an ideal gas at STP? J (1 mole ) 8.314 (273.15 K ) nRT ( mol K ) V = = = 0.00224 m3 = 22.4 liters p 1.01×105 Pa For any ideal gas Problem Solving with the Ideal Gas Law Useful facts and definitions: • Standard temperature and pressure (STP) T = 273 K (0°C) P = 1.00 atm = 1.013 × 105 N/m2 = 101.3 kPa • Volume of 1 mol of an ideal gas is 22.4 L • If the amount of gas does not change: p1 V1 p2 V 2 = [fixed n] T 1 T 2 • Always measure T in kelvins • P must be the absolute pressure Kinetic model of a monoatomic ideal gas Let us try to link this macroscopic model with a microscopic description of the system. The force exerted on the wall by the collision of one molecule is 2 Δ mv 2 mvx mvx F = = = Δ ℓ/ ℓ t 2 v x Then the force due to all molecules colliding with that wall is m F = N v 2 ℓ x 2 But the average squares of speed components are equal. So, vx = 2 (1/3)v 2 m v ⇒ F = N x ℓ 3 And the pressure on the wall is: 2 2 F 1 N m v 1 N m v p = = x = x A 3 A ℓ 3 V Kinetic model of a monoatomic ideal gas … continued 2 1 = 2 Rewrite last equation: pV N m vx 3 ( 2 ) But we wrote before: pV = NkT 1 3 Therefore: KE = m v 2 = kT molecule 2 2 1 3 and KE = N m v 2 = N kT gas ( 2 ) 2 ACT: Molecular speeds Two separate containers of gas are in thermal equilibrium with each other. One contains He and the other contains Ar. Which of the following statements is correct? 2 2 <x > is equivalent to writing x A. vHe v Ar 2 2 B. vHe v Ar 2 2 C. vHe v Ar Root-mean-square velocities The key quantity is <v 2> mean of squared velocities Take the square-root back to the units of velocity 2 12 3 vrms v mv kT 2rms 2 Example: 3 molecules at 400,500,600 ms-1 400 500 600 average 500 m/s 3 Not the same! 4002 500 2 600 2 rms 507 m/s 3 Distribution of Molecular Speeds These two graphs show the distribution of speeds of molecules in a gas, as derived by Maxwell. The most probable speed, vP, is not quite the same as the rms speed. As expected, the curves shift to the right with temperature. © 2014 Pearson Education, Inc. pT diagram Melting curve (solid/liquid transition) Critical point solid liquid Vapor pressure curve Triple (gas/liquid transition) point gas Sublimation curve (gas/solid transition) pT diagram for N2 At 1 atm, Tboiling = 77 K Tmelting = 63 K Triple point for N2: p = 0.011 atm, T = 63 K p solid liquid 1 atm demo Triple point gas DEMO: N2 triple point 63 K 77 K T pT diagram (water) For water, the pT diagram looks a little different. liquid solid gas other water pT diagram for water DEMO: p Regelation of ice Critical point solid liquid demo 1 atm gas T 0°C 100°C pT diagram for water DEMO: p Boiling water with ice Critical point solid liquid demo 1 atm gas T 0°C 100°C Triple point (water) Triple point for water: p = 610 Pa (0.006 atm), T = 273.16K p solid liquid Critical point Liquid water Triple does not exist 610 Pa point for p < 610 Pa!! gas T At triple point Tmelting = Tboiling Critical point (water) Critical point for water: 647K and 218 atm p Supercritical fluid Liquid and gas are indistinguishable beyond critical point solid liquid Critical point Triple At critical point, ρgas = ρliquid point gas T ACT: pT diagram for CO2 Which of the following states is NOT possible for CO2 at 100 atm? A. Liquid B. Boiling liquid C. Melting solid D. Solid E. All of the above are possible. Melting (at ~ -50°C) For a boiling transition, solid liquid pressure must be lower:. Boiling (at ~ -5°C) Triple point for CO2 has a pressure > 1 atm. At normal atmospheric pressure (1 atm), CO2 can only be solid or gas. “Dry Ice” Sublimation at T = -78.51°C.
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