Unformatted text preview: then that same pressure is being exerted by the walls of the container in keeping it inside (by Newton’s Third Law)  Atmospheric pressure is 101.325 kilopascals (kPa), also known as 1.01 bars, 1 atmosphere (atm) and 760 millimetres of mercury in a barometer (mmHg, or Torr) For any column (including the column of air that extends to the top of the atmosphere to give us atmospheric pressure), the density and pressure equations can be rearranged to give P = ρgh (h being the height of the column) The volume of one mole of an ideal gas at 0°C and 101.325 kPa is 22.4 L. If you forget, you can calculate it from PV = nRT. Ethan Newton & Barry Zhang for SOS Winter 2012 17 Conversely, if you forget R but remember 22.4, you can calculate R from this as well. You can also calculate R for other combinations of units. The laws of physics (kinematics, work and energy, momentum, etc) also apply to gases. The lecture notes go into great detail on this point.  The variables you use are different, though: kinetic energy is ek, speed is u, impulse is I  also, N is the number of molecules, and NA is Avogadro’s number  remember that ideal gas collisions are always perfectly elastic.  equations you should probably remember in addition to your physics equations are ēk = (3/2)RT = (3/2) PV 3RT urms =
M
 the latter one can also be expressed using Boltzmann’s constant, kB = R/NA Maxwell Boltzmann distribution €
 a probability distribution, indicating the percentage of gas molecules at any given speed  the curve starts at 0 m/s, increases to the maximum, then decreases asymptotically to zero as speed approaches infinity  as temperature increases, the peak becomes lower and the range of speeds becomes wider  as temperature decreases, the peak becomes higher and the range of speeds becomes narrower  the area under the curve always remains the same (for the same system)  different gases have different curves at the same temperature based on their molecular weight (what is the relationship?) um, the modal speed, is the speed space as it likes – there will be a mass exodus of molecules out of the liquid into the gaseous phase. − Since the vapour pressure above the liquid increa...
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This note was uploaded on 10/02/2012 for the course CHEM 112 taught by Professor Carran during the Winter '08 term at Queens University.
 Winter '08
 carran
 Chemistry, Atom, Quantum Chemistry, Stoichiometry

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