[Sis290025]

a. Predict the sign of  S for each of the following processes, which occur at constant temperature.      

I: Volume of 2 mol O2 (g) increases from 44 L to 52 L
II. Pressure of 2.0 mol O2 (g) increases from 1.0 atm to 1.2 atm

a.   I:  S= negative; II:  S= negative  
b.   I:  S= negative; II:  S= positive  
c.   I:  S= positive; II:  S= negative  
d.   I:  S= positive; II:  S= positive
   
I don't know if this is the right method, but I used n =PV/RT to see if the mol of gas increases, which increases disorder and entropy value. If the V or P rises, n rises.





b. For which of the following processes would the change in entropy be negative?

i. water boiling
ii. lead melting
iii. water condensing  

a.   i only  
b.   ii only  
c.   iii only  
d.   i and ii  
e.   i and iii


If S is negative, then more order is present, so the the process must yield a substance not in gaseous form. Condensation has liquid as product.

Thanks.    

[Sis290025]

Are my reasonings incorrect?

Thank again.

[Yggdrasil]

For a the answer is c.  In both cases n remains constant.  Rather, its only P and V which are changing.  In case I, V increases and P decreases.  In case II, P increases and V decreases.

Now, we can consider entropy as a measure of the "degrees of freedom" of a molecule.  By expanding the volume, the gas molecules have more room to move around and therefore the entropy is higher.  Conversely, decreasing volume constrains the freedom of movement of the molecules and decreases entropy.  Since temperature is constant, this is the primary effect which is causing the change in entropy during the expansion/compression of the gas.  Therefore, for the expansion in case I, the change in entropy is positive, while the change in entropy is negative for the compression in case II.

Your reasoning for b, however, is entirely correct.  Good job.