Energies of the diametric molecules of a gas, chemistry, Microeconomics

Energies of the diametric molecules of a gas, chemistry assignments

The analysis basis for treating these different types of motion can be seen by describing the motion of a diametric molecule of a gas. Here we describe the potential and kinetic energy components of a freely moving gas phase molecule treated as if it were a ball and spring system.

The only potential energy contribution arises from variations in the distance between the atoms of the molecule. If the variable intermolecular distance is represented by, then the energy U can be shown to be a function of r by writing U(r).

The kinetic energy depends on the motion of the two atoms of the molecule dx1/dt, dy1/dt and dz1/dt, respectively, of one of the atoms. The symbols x2y2 and z2 represent the velocity components of the second atom.

The total mechanical energy ε of the diatomic molecule is given by;

Separate translational, rotational and vibrational components of this energy can be recognized if a different coordinate system is introduced.

The center of mass of the molecules can be located by coordinates represented by X, Y and Z. the center of the mass, as is illustrated in two dimensions in fig. is related to the atomic pressure by the expressions:

ε = ½ m1 (x21 + y21 + z21) + ½ m2(x22 + y22 +z22) + U (r)

(m1 + m2)X = m1x1 + m2x2

(m1 +m2)Y = m1y1 + m2y2

(m1 + m2)Z = m1z1 +m2z
2

The orientation of the molecule is expressed by the polar angular coordinate's θ and Ø. These angular coordinates and the internuclear distance r lead to the relations:

x2 - x1 = r sinθ cos∅, y2 - y1 = sinθ cosØ, z2 - z1 = r cosθ

These relations can be used to eliminate the coordinates for one or the other of the two atoms. We can, for example, use the first to write x2 = x1 +r sinθ cos Ø. Substitution in the first step of eq.  eliminates the x2term. This procedure leads to:

x1 = X - m2/m1 + m2 r sinθ cos Ø 

y1 = Y- m2/m1 + m2 r sinθ cos Ø 

z1 = Z - m2/m1 + mr cos θ 

and, x2 = X - m2/m1 + m2 r sinθ cos Ø 

y2 = Y - m2/m1 + m2 r sinθ cos Ø 

z2 = Z - m2/m1 + m2 r cos θ 


the derivates of expressions with respect to time can be taken if we recognize that X, y, X, r, θ and Ø are all time dependent. The results for x1, y1, z1 and x2, y2z2 can be substituted to give, after rearrangement:

ε = ½ (m1 + m2) (X2 + Y2 +Z2) + ½ m1m2/m1 +m2 [r2 + r2 θ2 + r2 (sin2θ) Ø2] +U (r) 

Posted Date: 2/15/2012 10:01:55 AM | Location : United States







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