Sapling W5/6 #7


Moderators: Chem_Mod, Chem_Admin

Jillian Labador 3E
Posts: 118
Joined: Wed Sep 30, 2020 9:42 pm

Sapling W5/6 #7

Postby Jillian Labador 3E » Sun Feb 14, 2021 11:44 pm

Hi ! For the second part of the question, it asked "Based on the calculated ΔSvap values, arrange the liquid samples in order from most ordered to least ordered" and the answer was:

most ordered
CH3OH
C3H6O
C5H12
least ordered

The solution stated that the most ordered liquid is the one with the largest ΔSvap. Can someone explain why? I thought it was the other way around.

Benjamin_Hugh_3F
Posts: 105
Joined: Wed Sep 30, 2020 9:33 pm
Been upvoted: 1 time

Re: Sapling W5/6 #7

Postby Benjamin_Hugh_3F » Mon Feb 15, 2021 12:28 am

A larger disorder correlates to a higher ΔSvap, which means the molecules are more ordered in the liquid state. Thus high ΔSvap means molecules ordered liquid state and low ΔSvap means molecules are less ordered in liquid state.

edward_brodell_2I
Posts: 92
Joined: Wed Sep 30, 2020 9:59 pm

Re: Sapling W5/6 #7

Postby edward_brodell_2I » Mon Feb 22, 2021 8:20 pm

Entropy is a measure of the possible positions (degeneracies) that the atoms can take. So if there is more temperature (therefore, more random thermal motion), then the system can take on more possible states and so it would be harder to maintain or achieve a particular state since there are more positional states for atoms. In terms of the entropy changes, since there is the greatest increase of possible positions that the atoms in a molecule gain due to the entropy change, then the liquid must have been the most ordered since it is gaining the most change in random movement of atoms. If a substance already had many possible positions or states, then the entropy change would be smaller since the atoms already have many different states.


Return to “Concepts & Calculations Using Second Law of Thermodynamics”

Who is online

Users browsing this forum: No registered users and 6 guests