Vaporization and change in internal energy

[songoku]

Homework Statement

Suppose 1 g of water vaporizes isobarically at atmosphere pressure (1.013 x 10⁵ Pa). Its volume in the liquid state is Vi = 1 cm³, and its volume in the vapor state is Vf = 1671 cm³. Find the change in internal energy

Homework Equations

[tex]\Delta U = \Delta Q - \Delta W[/tex]

[tex]\Delta W = p*\Delta V[/tex]

The Attempt at a Solution

I want to ask about [tex]\Delta V[/tex].
[tex]\Delta V = V_2-V_1[/tex]

V₂ = Vf, and what is V₁ ? Is it Vi or is it zero because volume of vapor is zero initially?

Thanks

 

[tiny-tim]

Hi songoku!

(have a del: ∆ )

Suppose 1 g of water vaporizes isobarically at atmosphere pressure (1.013 x 10⁵ Pa). Its volume in the liquid state is Vi = 1 cm³, and its volume in the vapor state is Vf = 1671 cm³. Find the change in internal energy
…
V₂ = Vf, and what is V₁ ? Is it Vi or is it zero because volume of vapor is zero initially?

It's V_i.

 

[songoku]

Hi tiny-tim!

Why is it not zero? I think we have to consider the vapor state, excluding the liquid state. Does the first law of thermodynamics only apply for gas?

Thanks

 

[Redbelly98]

The first law of thermodynamics applies to everything, not just gasses.

 

[songoku]

I encountered a very similar question.

Consider 100 g (100 cm³) of a liquid evaporating at constant pressure of 100 kPa to vapor of volume 0.167 m³. Assuming that the latent heat of vaporization of the liquid is 2.26 MJ kg^-1 K^-1 and the vapor behaves like an ideal gas, find the change internal energy.

On the manual, it is written :

note that the volume of gas is zero initially, so ∆V = 0.167 - 0

Is the manual wrong or am I missing something?

Thanks

 

[Redbelly98]

That's weird.

It makes very little difference in the result, but I was pretty sure you should account for the original volume of the liquid -- if the accuracy of the given numbers warrants it.

 

[songoku]

ok

Thanks for your help, tiny-tim and Redbelly98 !