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a state has an energy e=76.5e-21 and a degeneracy 15. What is the free energy of the state at 25 degrees celsius

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5 votes

answer !!!

so basically:

To calculate the free energy of a state at 25 degrees Celsius, we can use the formula:

F = E - T*S

Where:

F is the free energy

E is the energy

T is the temperature in Kelvin

S is the entropy

Given:

Energy (E) = 76.5e-21

Degeneracy = 15

Temperature (T) = 25 degrees Celsius

To convert the temperature from Celsius to Kelvin, we add 273.15 to the Celsius temperature:

T = 25 + 273.15 = 298.15 K

To calculate the entropy (S), we can use the formula:

S = k * ln(D)

Where:

S is the entropy

k is Boltzmann's constant (approximately 1.38e-23 J/K)

D is the degeneracy

Substituting the given values into the formula, we have:

S = 1.38e-23 * ln(15)

Calculating the natural logarithm of 15, we find:

S ≈ 1.38e-23 * 2.708

S ≈ 3.729e-23 J/K

Now we can calculate the free energy (F):

F = 76.5e-21 - 298.15 * 3.729e-23

Calculating the product of the temperature and entropy, we find:

F ≈ 76.5e-21 - 1.113e-20

F ≈ -1.035e-20 J

Therefore, the free energy of the state at 25 degrees Celsius is approximately -1.035e-20 Joules.

im not exactly sure if this is right but i hope it at least helps - aydn

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