Athree level system has energies ε0=0.00j, ε1=5.00x10-21j, and ε2=10.0x10-21j. a. using the boltzmann distribution law, calcuate the molar internal energy u at t=10.0k. hint: this means you have to calculate three probabilites at t=10k and use them to calculate the average energy per particle and then calculate the energy of a mole of particles. b. again using the bolztmann distribution law and the same approach as in part a, calculate the molar internal enerrgy u at t=1000kc. using you results from parts a and b, calculate the change in the molar internal energy δu when the temperature changes from t=10k to t=1000k. d. using the bolzmann distribution law probabilities calculated in part a, calculate the molar entropy at t=10k. e. calculate the change in molar entropy when the temperature of this 3 level system changes from t=10k to t=1000k.

A three level system has energies ε0=0.00J, ε1=5.00x10-21J, and ε2=10.0x10-21J. a. Using the Boltzmann Distribution Law, calcuate the molar internal energy U at T=10.0°C. Hint: This means you have to calculate three probabilites at T=10K and use them to calculate the average energy per particle and then calculate the energy of a mole of particles.b. Again using the Bolztmann Distribution Law and the same approach as in part A, calculate the molar internal enerrgy U at T=1000°C c. Using you results from parts A and B, calculate the change in the molar internal energy ΔU when the temperature changes from T=10°C to T=1000°C.d. Using the Bolzmann Distribution Law probabilities calculated in part A, calculate the molar entropy at T=10°C.e. Calculate the change in molar entropy when the temperature of this 3 level system changes from T=10°C to T = 1000°C

A). 960.9J/mol

B). 2447.31J/mol

C). 1486.41J/mol

D). 577.706JK-1mol-1

E). 576.233 JK-1mol-1

F). 1.473JK-1mol-1

Explanation:

partition function q equals

(a) q = 1+e(−ε1/kT) + γe(−2ε2/kT) .

 As such substituting the values for the energies at the different states we have where putγ= 1

q = 1 + 0.2779+7.726 = 1.3552

From where we have populations, or probabilities, P1, P2 , P3

P1 = 1/q= 1/1.3552= 0.7378

P2 = e−ε1/kT/q = 0.2051

And  P3 = e−2ε2/kT/q = 0.05701

The Average energy per particle can be found as

[ε0] = 0P0 + ε0Pi + ε1 P2

= 0 + 5×10-21 ×0.2051 + 10×10-21× 0.05701 = 1.5956×10-21J

1.5956×10-21J×6.02×1023 particles= 960.9J/mol

B). at 1000 C we have also

q = 1+e(−ε1/kT) + e(−2ε2/kT)

From where q = 1+0.7523+0.5659 = 2.31825

From where P1 = 0.43136, P2 = 0.3245 and P3 = 0.244

And the average energy is

[ε0] = 0P0 + ε0Pi + ε1 P2

= 4.0638×10-21J hence U = 4.0638×10-21J × 6.02×1023 particles = 2447.31J/mol

C). Change in molar internal energy between 10 and 1000C =

2447.31J/mol-960.9J/mol = 1486.41J/mol

D).  at 10C we have

S/n = R×lnq + ∆U/(n×T)

∆U/n = 960.9J/mol hence

S/n = (8.314JK-1mol-1) ln(1030) + 960.9J/(283 mol K)

 = S/n = 577.706JK-1mol-1

E). At 1000C we have

S/n = (8.314JK-1mol-1) ln(1030) + 2447.31J/(1273 mol K) = 576.233 JK-1mol-1

F). Difference = 577.706JK-1mol-1 - 576.233 JK-1mol-1 = 1.473JK-1mol-1

a electronegativity decreases beacuse the size of nucleus is bigger.