number 4 please? I think the trick is that the DNA bases can be treated as point

Question

number 4 please? I think the trick is that the DNA bases can be treated as point particles and so I found the energy using E=(q1*q2)/(4*pi*vacuum constant*r) where r is the distance between the particles. if this is correct, then I am having trouble finding the change in entropy from this.

thanks in advance, and I hope you have a great day!

Calculate the interaction energy between two freely rotating molecules at 300K, 1 nm apart, each with a dipole moment of 1 Debye. Answers in Joules and ev please. (8 points) Calculate the critical micelle concentration for a methane molecule (ra0.2nm) in water (interface energy with water 50ml/m2) at 350K. (4 points) 4. Estimate the minimum amount of bonding energy to hold two DNA bases together to overcome the Coulomb repulsion between phosphates (effective charge 0.1 e each, separation 2nm) and the loss of entropy from pairing the bases. List each energy separately (in Joules). For the entropy, assume that prior to incorporation into the bonded pair, the bases are completely free (T 300K). (10 points)

Explanation / Answer

Force between two particles can be given by Coulomb's interaction force

Q= (1/4*pi*absilon)*(q1.q2)/r2 where q1, q2 are the charge on the two particles and r is the separation between them

Q= (9*109) * (0.1*1.6*10-19)2/(2*10-9)2

= 5.76 * 10-13 J

Loss in entropy, dS = dQ/T

= (5.76 * 10-13)/300 J/K

= 1.92 * 10-15   J/K

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