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Is it

A) J
B) J·mol–1
C) J·mol–1·K–1
D) J·K·mol–1

I think its C, but im not sure if the mol are in the denom or not..

2006-10-23 04:40:54 · 5 answers · asked by axcryingxshame 1 in Science & Mathematics Chemistry

5 answers

Your problem is apparently looking for the "Molar entropy" which is different than "entropy."

Molar entropy would be answer C,

joule per mole kelvin: J/mol K



Aloha

2006-10-23 04:47:43 · answer #1 · answered by Anonymous · 0 0

The definition of entropy, S, is Q/T, where Q is heat and T is a temperature. The units of Q are joules, and the units of T are kelvins, making the units of entropy J/K, or J*K^-1. What you have available is choice C, which is actually entropy per mole, a form of specific entropy. But you are right to think that C is correct.

2006-10-23 11:47:02 · answer #2 · answered by DavidK93 7 · 1 0

One way to define the quantity "entropy" is to do it in terms of the multiplicity.

Multiplicity = W
Entropy = k lnW


where k is Boltzmann's constant. The k is included as part of the historical definition of entropy and gives the units Joule/Kelvin in the SI system of units. The logarithm is used to make the defined entropy of reasonable size. The multiplicity for ordinary collections of matter are on the order of Avogadro's number, so using the logarithm of the multiplicity is convenient.

For a system of a large number of particles, like a mole of atoms, the most probable state will be overwhelmingly probable. You can with confidence expect that the system at equilibrium will be found in the state of highest multiplicity since fluctuations from that state will usually be too small to measure. As a large system approaches equilibrium, its multiplicity (entropy) tends to increase. This is a way of stating the second law of thermodynamics.

2006-10-23 11:50:03 · answer #3 · answered by Brite Tiger 6 · 0 1

Entropy in SI units is measured in joules per degree kelvin, or J.K-1. "C" is the molar entropy: joules per mol per degree K.

2006-10-23 11:49:58 · answer #4 · answered by JJ 7 · 0 1

(J·K−1).

2006-10-23 11:43:16 · answer #5 · answered by Anonymous · 0 1

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