Tangential Velocity Homework: Attractive Energy & Force Calc.

[Priscilla]

Homework Statement

A carbohydrate gel is being centrifuged to remove excess physisorbed water. Assume that the magnitude of the attractive energy between the water molecules and the gel is given by 3.63 kJ/mol of molecules of water, with the water molecules being separated from the surface of the gel molecules by 0.4 nm.
a)Calculate the attractive energy per molecule of water, and the attractive force between one molecule of water and one gel molecule.
b)Assuming that the gel in the centrifuge has a radius of curvature of 0.5 m when the centrifuge rotates, find the minimum tangential velocity with which the centrifuge needs to rotate in order for water molecules to just begin to separate from the gel molecules, at a separation of 0.4 nm

Homework Equations

N_A = 6.02*10^23 molecule/mol
E=fr
F_c = mv^2/r

The Attempt at a Solution

The attractive energy per molecule of water
E = 3.63kJ/mol of molecules of water
E = 3.63kJ/mol / 6.02*10^23 molecule/mol = 6.03*10^-24 kJ/molecule of H2O

The attractive force between one molecule of water and one gel molecule
E = fr
f = E/r = (6.03*10^-24 kJ/molecule) / 0.4*10^-9m = 1.51*10^-14 N/molecule of gel

The minimum tangential velocity
F_c = mv^2/r
F = 1.51*10^-14 N r = 0.5m
I know I can use this equation to find the velocity, but I don't know the mass. How can I find the mass?

 

[mark726]

To find the mass, we can use the formula for the centripetal force, where F_c = m*v^2/r. We know the force (1.51*10^-14 N), the radius (0.5 m), and we can assume a mass of one water molecule (1.67*10^-27 kg). Plugging these values into the equation, we get:

1.51*10^-14 N = (1.67*10^-27 kg) * v^2 / 0.5 m

Solving for v, we get a minimum tangential velocity of 0.0036 m/s. This is the minimum velocity needed for the water molecules to just begin to separate from the gel molecules at a separation of 0.4 nm.