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### Changes in both volume and temp

Posted: Sat Feb 11, 2017 7:06 pm
If we are given a problem that provides us with both an initial volume and temp and a final volume and temp do we have to calculate deltaS for both of these or will 1 calculation of deltaS suffice? I am confused on if we do both S=nCln(T2/T1) and S=nCln(V2/V1)? Thank you very much in advance

### Re: Changes in both volume and temp  [ENDORSED]

Posted: Sat Feb 11, 2017 7:25 pm
One is enough. You can do the other to double check your work as the value should be the same.

The mathematical rationale behind this is based on the ideal gas law: PV = nRT from which you can see that volume and temperature are directly proportional, so the ratios should be the same.

### Re: Changes in both volume and temp

Posted: Tue Feb 14, 2017 5:52 pm
I am confused by Chem_Mod's answer. I have been under the impression that we need to use BOTH DeltaS = nClnT2/T1 AND DeltaS = nRlnV2/V1 if both the temperature and volume change.

For example, if initial temperature = 10 K, initial volume = 5 L, final temperature = 20 K, and final volume = 40 L, then:

Step 1 = At 10K, the volume changes from 5 L to 40 L.
Step 2 = At 40 L, the temperature changes from 10K to 20K.

In Midterm 2015, 4B does what I stated above.

Can someone correct me on why I disagree with Chem_Mod?

### Re: Changes in both volume and temp

Posted: Tue Feb 14, 2017 6:41 pm
The assumption for the first question was that it was at constant pressure, in which case all three methods work:
is the same as when pressure is constant (also assuming ideal gas), and since , this is the same as . If pressure is not constant, the process must be broken up into two steps, one at constant volume using , and the other at constant temperature using .

### Re: Changes in both volume and temp

Posted: Wed Feb 15, 2017 5:24 pm
How would you be able to tell whether or not a reaction occurs at a constsnt temperature?

### Re: Changes in both volume and temp

Posted: Sun Feb 26, 2017 10:31 pm
I think the problem would state that the reaction will occur at constant temperature so that we know which equation to use.