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A Little Observation

A Little Observation. Consider 1 mole of water molecules. H 2 O has a molecular weight of 18 gm/mole. So, 6.02x10 23 molecules weigh 18 gm. But, 18 gm of liquid water occupies 18 ml volume (18 cm 3 ). Liquid water is pretty incompressible. So we guess that the

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A Little Observation

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  1. A Little Observation Consider 1 mole of water molecules. H2O has a molecular weight of 18 gm/mole. So, 6.02x1023 molecules weigh 18 gm But, 18 gm of liquid water occupies 18 ml volume (18 cm3) Liquid water is pretty incompressible. So we guess that the water molecules in the liquid are literally in contact with each other. No significant space in between.

  2. So the actual volume of one water molecule must be roughly 18 ml/6.02x1023= 3x10-23 ml Contrast this with the volume occupied by a water molecule in the gas phase. To find this number, treat water as an ideal gas at 300 K and use the ideal gas law to compute the volume: pV=nRT with n=1 mole, p= 1 atm. and R=0.082 l-atm/mole-deg V=(1)(.082)(300)/(1)= 24.6 liters Or, the volume occupied per molecule is 24,600ml/6.023x1023= 4.1x10-20 ml Compare! Thus the ratio of the volume occupied by an ideal gas molecule to its actual volume is 24600/18=1400 !!

  3. This leads us to conclude that molecules in an ideal gas must be far apart, “rarely” bumping into each other. The volume of an actual molecule is tiny by comparison to the volume occupied at 1 atmosphere and 300K in the gas phase. The picture we walk away with for the gaseous state of a molecule in an ideal gas is one of huge empty spaces between molecules with “rare” collisions. This will allow us to develop a simple MODEL of the gaseous state which provides remarkable insight into the properties of molecules and matter. This MODEL is called the Kinetic Theory of Gases.

  4.  v = +10 cm / s, c = 10 cm / s v = -10 cm / s, c = 10 cm / s -x +x 1) Particle velocity includes both the speed ( c ) of a particle (cm/s) and its direction. In one dimension: 0 2) Particle Momentum is P = mv (not mc). In one dimension:   mgm at c = 10 cm/s m gm at c = 10 cm/s   P = - 10m gm-cm/s P = + 10m gm-cm/s -x +x 0 Kinetic Theory Preliminaries 3) Change in Momentum for an elastic collision: An elastic collision is one where speed is the same before and after the collision:

  5. v =+10cm/s -x +x { v = -10cm/s c=10cm/s c=10cm/s Wall

  6.  Pinitial = +10m(gm-cm/s) = mv = mc   Pfinal = -10m(gm-cm/s) = mv = -mc    ∆P = Pf - Pi = -mc - (mc) = -2mc [∆ is the symbol for “change”]    ∆Pwall = +2mc ∆Pw + ∆Pm = 0 4) Conservation of Momentum: what particle loses, wall must gain

  7. 5) Acceleration a is  (velocity) /  (time). a like v has direction.       a = ∆v / ∆t ( a = dv / dt )   6) Force = F = ma     F = ma = m∆v / ∆t = ∆P / ∆t   F = dP / dt Force is change in momentum with time:

  8. Kinetic Theory of Gases Assumptions 1) Particles are point mass atoms (volume = 0) 2) No attractive forces between atoms. Behave independently except for brief moments of collision. Model System A box of volume V with N atoms of mass m all moving with the same speed c. V, N, m, c We wish to calculate the pressure exerted by the gas on the walls of the box:

  9. Typical Path for a gas atom or molecule in a box. A A Force of atom impinging on wall creates pressure that we can measure [pV = nRT].

  10.   F = ma = m( ∆v )/∆t (∆P)/∆t=(Change in momentum) / (change in time)  Pressure  Force / unit area Thus, we need to find force exerted by atoms on the the wall of the box. Let’s try to calculate the force exerted by the gas on a segment of the box wall having area A. To do this we will make one more simplifying assumption: We assume that all atoms move either along the x, y, or z axes but not at any angle to these axes! (This is a silly assumption and, as we shall see later, causes some errors that we must correct.)

  11. Vectors and Vector Components: The Movie Z Y X

  12. Wall Collisions  Vessel wall Vessel wall Approach mv mvy -mvx mvx mvy Recoil mv Before Collision with wall After Collision with wall A more accurate view of elastic collisions with a wall. The component of momentum perpendicular to the wall reverses sign, from mvx to -mvx. The component parallel to the wall, mvy, is unchanged. Total momentum is shown by a red arrow. Although the direction of v is changed by the collision, its length is not. Molecular speeds are unaffected by elastic wall collisions.

  13. Y X Z ct wall Now construct an imaginary collision cylinder as follows: Area of cylinder is A, length is ct where c is atom speed, t is some arbitrary time: A This cylinder contains all the atoms which will strike A in a time t (It also contains quite a few atoms that will not collide with the wall during t). Remember that all atoms have been assumed to move perpendicular to the wall.

  14.  F = ( m∆v )/∆t  Fatom/atom = (-2mc)/∆t This is the force exerted ON an atom due to a single collision.  Fwall/atom = (+2mc)/∆t The force on the wall due to a collision by a single atom can be computed as follows: Since the momentum change for the wall is the negative of that for the atom:

  15. One Particle Momentum Change for Elastic Wall Collision  Wall m m m∆v = m(-c - (+ c)) = -2mc v = - c v = + c

  16. 2mc = (∆Pwall)/(impact) Then: [(Momentum change) / sec] = (∆ Pwall / impact) • ( impacts / sec) = (∆ Pwall / sec)     F = (2mc) • I = (∆ Pwall / sec)  To obtain F all we need now is I which is relatively easy to calculate: Our problem now is to determine ∆t. There is no easy way to do this so we resort to a trick: Let I = # of impacts of atoms with wall / sec.

  17. A ct  Fwall = [(2mc)][(1 / 6)(N / V)(Ac)] = (1 / 3)(N / V)mc2A N / V = (#atoms) / cm3 Calculating I A(ct) = volume of collision cylinder: Total atoms in collision cylinder = (N / V) (Act) Since 1/3 of the atoms move along x, 1/3 move along y, 1/3 move along z and one half of the atoms moving along any direction are moving towards the wall while one half move away from the wall, the # of atoms in the collision cylinder which will strike the wall in a time t is: (1/3)(1/2)(N / V) (Act) = (1/6) (N / V) (Act) Directions/axis # of axes # impacts / sec = I = ((1/6)(N / V)(Ac t)) / t = [(1/6)(N / V)(Ac)]

  18. P = F / A = [(1 / 3)(N / V)(mc2A)] / A The pressure is the force / unit area or P = (1/3) (N / V)mc2 or  PV = (2/3) N [(1/2) mc2] Where N is the total number of atoms in the box and (1/2) mc2 is the kinetic energy of one atom. Experimentally PV = nRT suggesting that at constant T, the kinetic energy of an atom is constant. Let N0 = Avogadro’s #; n = # moles in V = N / N0 PV = N (RT / N0) = (2/3) N [(1/2) mc2] or 

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