Enthelpy change of a vaporization rxn

AI Thread Summary
The standard enthalpy of vaporization for the liquid is 32 KJ/mol, and when 0.75 mol is vaporized at 260 K and 765 Torr, q equals delta H due to constant pressure conditions. The discussion highlights that work is not zero because the volume increases, indicating that the system does work on its surroundings. This distinction emphasizes the difference between energy and enthalpy, as enthalpy accounts for volume changes during processes like vaporization. The calculation of ∆H involves adjusting the enthalpy of vaporization to the number of moles. Understanding these concepts is crucial for solving related thermodynamic problems.
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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.



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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?
 
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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.
 
I have a similar question, just wondering how you solved ∆H? I can do the rest.

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