M1M1 M1M1 m1m1 m1m1 Case 1 Case 2 M2M2 m2m2 +d -d KE trans Speed PE gravity Height KE trans Speed PE gravity Height ½ M 1 (v f 2 -0) +M 1 g(h f -0)+ ½ m 1 (v f 2 -0) + m 1 g(h f -0)=0 ½ M 1 v f 2 +M 1 g(-d)+ ½ m 1 v f 2 + m 1 g(d)=0 (M 1 +m 1 )½v f 2 + (m 1 -M 1 )gd =0 Combining PE and KE terms PE’s are the same for both systems (mass difference is the same) So KE’s must be the same for both systems But… M+m is bigger for case 1, therefore: vf must be smaller to make up for it! M1M1 M2M2 m2m2 m1m1
Ideal Gas In Intro Chemistry we always dealt with ‘Ideal’ gasses. What does that actually mean? Ideal gases: – Have no intermolecular forces – Have perfectly elastic collisions with each other (and the sides of containers)
What was the point of the N 2 Activity? What did we calculate? Spacing of atoms is about 10σ. At what point of the pair-wise potential do atoms/molecules have zero PE and Zero force? 3σ! What do we take away from this? The ideal gas approximation is useful for gases!
Intro Particle Model of Matter A graphical representation of the energies associated with particles
We know the shape… but what exactly is this a graph of? A.The potential energy of one atom with respect to a system of particles. B.The potential energy of a system (many particles) C.The potential energy of one particle with respect to another particle D.The total energy of one particle with respect to a system of atoms E.The total energy of one atom with respect to another
Remember the Anchor But Cassandra when is one particle ever ‘anchored’ in space? Good question! It’s not, but our graph is always drawn with respect to one particle at the origin, even if the origin is moving
Energy r (atomic diameters) r is the atomic diameter roro is the well depth r o is the equilibrium separation Potential Energy between two atoms “pair-wise potential” a.k.a. Lennard-Jones Potential pair-wise ~ 10 -21 J ~ 10 -10 m = 1Å Do not need to memorize
Forces and the Potential RepulsiveAttractive Force = -d(PE)/dx Or, negative change in y over change in x Force has a magnitude of slope, and the direction of decreasing PE!
If the curve only tells us about PE, how do we find KE and Etot?
E tot Etot = KE + PE -3ε = KE + -3ε -3ε = KE + -4ε KE = 0 KE = 1ε -3ε = KE + -7εKE = 4ε -3ε = KE + -8ε KE = 5ε -3ε = KE + -7ε KE = 4ε -3ε = KE + -4εKE = 1ε -3ε = KE + -1εKE = -1ε KE can’t be negative!!!!!
E tot Turning Points Where the Etot intersects the PE curve, there are ‘turning Points.’ The particle oscilates between These two points.
How much work does it take to move one particle from rest at equilibrium (1.12σ), to 3σ with a minute (negligible but non zero) velocity? A.1ε B.3ε C.-1ε D.2.88σ E.Impossible to tell if
Same idea as before: Initial: at 1.12σ, v=0 PE + KE = Etot -1ε + 0 = -1ε Now what? Is this a closed system? NO! Adding energy: Final: at 3σ, v~0 So new Etot = 0 Must add 1ε to get there.
OK let’s draw an Energy System Diagram: PE pair- wise System: Two Particles, one bond Initial: v=0, r=1.12σ Final: v~0 r=3σ Wait! We don’t have an equation for PE pair-wise! It’s ok, we have something better… a graph! Work ΔPE = Work PE f – PE i = Work 0ε – (-1ε) = WorkWork = 1ε Energy Added i f
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