Enthalpy change of a vaporization rxn

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Homework Statement



A certain liquid has a standard enthalpy of vaporization of 32 KJ/mol. Calculate q, w, delta H and delta U when 0.75 mol is vaporized at 260 K and 765 Torr.



Homework Equations





The Attempt at a Solution



I was able to get delta H, but the solution states that q is equal to delta H. If delta H is equal to q at constant pressure, I don't know how I'm supposed to know when to make that assumption. Also, why does work not equal zero if the pressure is constant, is this additional work?
 

Answers and Replies

  • #2
Mapes
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The amount of work is nonzero because the volume increases and the system therefore does work on its surroundings. This is the big difference between energy and enthalpy; enthalpy takes into account that a system's volume tends to vary during processes such as heating and vaporization.
 
  • #3
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I have a similar question, just wondering how you solved ∆H? I can do the rest.

Thanks
 
  • #4
Mapes
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Hi hinglis, welcome to PF. ∆H is the enthalpy of vaporization, adjusted to the number of moles in the system.
 

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