Consider a disordered crystal of monodeuteriomethane in which each tetrahedral C

Question

Consider a disordered crystal of monodeuteriomethane in which each tetrahedral CH3D molecule is oriented randomly in one of four possible ways. Part A Use Boltzmann's formula to calculate the entropy of the disordered state of the crystal if the crystal contains 13 molecules. Part B Use Boltzmann's formula to calculate the entropy of the disordered state of the crystal if the crystal contains 130 molecules. Part C Use Boltzmann's formula to calculate the entropy of the disordered state of the crystal if the crystal contains 1 molof molecules. Part D What is the entropy of the crystal if the CD bond of each of the CH3D molecules points in the same direction? Crystal contains 13 molecules. Part E What is the entropy of the crystal if the CD bond of each of the CH3D molecules points in the same direction? Crystal contains 130 molecules. Part F What is the entropy of the crystal if the CD bond of each of the CH3D molecules points in the same direction? Crystal contains 1 mol of molecules.

Explanation / Answer

A] Boltzmann's entropy formula is:

S = k*ln(w)

where k is Boltzmann's constant ( the universal gas constant divided by Avogadro's number, R/No) = 1.38*10^-23 J/K, and w is the number of "microscopic" (i.e., atomic and molecular states) that give rise to the same "macroscopic" state (i.e., observed energy state).

In this case each molecule can exist in 6 possible orientations,

S = k*ln(w) = (R/No)*ln(6^13)

Remember that ln(x^y) = y*ln(x), so:

S = [1.38*10^-23]*13 ln4 = 2.48*10^-22 J/K

B] same formula

S= k*ln (w) = k* ln (4^130) = 2.48 *10^-21 J/K

C] S = k* ln (4^(6*10^23)) = 19.79 J/K

For part D , part E , part F

since it is oriented in one direction only

Then , S= k* ln (w) =k* ln (1^x) = 0 ; since 1^x = 1 and ln1 =0 THe entropy of the crystal becomes zero in all D , E and F parts .

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