### 9.35

Posted:

**Wed Feb 13, 2019 11:42 am**Why does A, the monatomic gas have a larger enthalpy change than a diatomic gas?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=132&t=42182

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Posted: **Wed Feb 13, 2019 11:42 am**

Why does A, the monatomic gas have a larger enthalpy change than a diatomic gas?

Posted: **Wed Feb 13, 2019 3:12 pm**

Container A has a greater number of particles than the other containers. Therefore, the entropy change is greater. In comparing monatomic with diatomic, there will be more of the monatomic particles because it needs twice as many particles to equal one diatomic particle.

Posted: **Sat Feb 16, 2019 10:22 pm**

Here's a more detailed explanation taken from Lavelle's Chem 14B Solution Manual Errors 6th Edition pdf:

"Students are asked to rank the delta S for a series of gases during a temperature change.

∆S = nCv ln(T2/T1) for an isochoric process Gas A (1.00 mol monatomic ideal):

∆S = nCv ln(T2/T1) = (1.00 moles)(3R/2)ln(T2/T1) and because all undergo the same temperature change, delta S is essentially 3R/2.

Gas B (0.5 mol diatomic, no vibrational degrees of freedom):

∆S = nCv ln(T2/T1) = (0.5 moles)(5R/2)ln(T2/T1) and because all undergo the same

temperature change, delta S is essentially 5R/4.

Gas C (0.5 mol diatomic, 1 vibrational degree of freedom):

∆S = nCv ln(T2/T1) = (0.5 moles)(3R)ln(T2/T1) and because all undergo the same

temperature change, delta S is essentially 3R/2.

The answer provided (B < C < A) is incorrect. The answer should be B < (C = A)."

"Students are asked to rank the delta S for a series of gases during a temperature change.

∆S = nCv ln(T2/T1) for an isochoric process Gas A (1.00 mol monatomic ideal):

∆S = nCv ln(T2/T1) = (1.00 moles)(3R/2)ln(T2/T1) and because all undergo the same temperature change, delta S is essentially 3R/2.

Gas B (0.5 mol diatomic, no vibrational degrees of freedom):

∆S = nCv ln(T2/T1) = (0.5 moles)(5R/2)ln(T2/T1) and because all undergo the same

temperature change, delta S is essentially 5R/4.

Gas C (0.5 mol diatomic, 1 vibrational degree of freedom):

∆S = nCv ln(T2/T1) = (0.5 moles)(3R)ln(T2/T1) and because all undergo the same

temperature change, delta S is essentially 3R/2.

The answer provided (B < C < A) is incorrect. The answer should be B < (C = A)."