Enthalpy change when vaporizing 0.75 mol of liquid at 260 K

[yungwun22]

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?

 

[Mapes]

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.

 

[hinglis]

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

Thanks

 

[Mapes]

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