Page 1 of 1

### 9.19

Posted: Sun Jan 28, 2018 10:09 pm
For problem 9.19, why do we also cool the system? Aren't we just find entropy change of vaporization?

### Re: 9.19

Posted: Sun Jan 28, 2018 10:31 pm
can someone also explain this to me?

### Re: 9.19

Posted: Sun Jan 28, 2018 10:58 pm
Do you have to cool it to start backwards from 85 degrees C, since we are given the standard entropy of vaporization at 100 degrees C?

### Re: 9.19

Posted: Mon Jan 29, 2018 8:58 pm
this problem is similar to example 9.6 on page 329, you have to add up the entropies from the three reversible steps of heating the water to 100 deg C, phase changing, then bringing it back down to 85 degrees C. All these added will be the same entropy of vaporizing water at 85 deg C in one irreversible step .

1) So you basically start by calculating the the entropy change that comes with heating the water using deltaS=(C)ln(T2/T1)). 75.3(ln(373/358))=3.09 J*K^-1*mol^-1 - molar heat capacity and temp are given.
2) Now we vaporize the water: deltaSvap=109.0 J*K^-1*mol^-1 -this is given in the problem
3) we cool it now. deltaS=(Cp,m)ln(T2/T1)) Cp,m is 4R from the textbook. 4*8.314*ln(358/373)=-1.37 J*K^-1*mol^-1

Add them up to get 111 J*K^-1*mol^-1

### Re: 9.19

Posted: Thu Feb 01, 2018 8:28 pm
For the third part of the example in the book, How would we know to use Cp,m= 4R? And why didn't they use that value to calculate the step 1 of heating the liquid acetone?