Change in Internal Energy of 1kg Water to Steam?

[Lagraaaange]

Homework Statement

Hvap = 22.6E5 J/Kg. R=8.315E3 Jkmol^-1K^-1. 1kmol water = 18kg
Keeping in mind that latent heats are enthalpies, what is the change in internal energy when 1kg water is converted to steam at pressure of 1atm and temperature of 100C.

Homework Equations

dU = dQ - PdV.

The Attempt at a Solution

dQ = mHvap
PdV = 1atm (Vfinal - Vinitial) Vfinal >>Vinitial -> PdV = 1atm(Vfinal).
Assume ideal gas-> V = nRT/P =18kilomoles * R*373 / 1atm?

 

[SteamKing]

Is there an organized question here, or is it just stream of consciousness which you are writing down?

 

[Chestermiller]

You have 1 kg of liquid water in a closed container with a piston at 100 C and 1 atm. You add heat until all the water has vaporized to water vapor at 100 C and 1 atm (holding the pressure constant). What is the change in enthalpy ΔH for the water?

Chet

 

[Lagraaaange]

You have 1 kg of liquid water in a closed container with a piston at 100 C and 1 atm. You add heat until all the water has vaporized to water vapor at 100 C and 1 atm (holding the pressure constant). What is the change in enthalpy ΔH for the water?

Chet

The H is given as the heat of vaporization which is the change in enthalpy. How do I get internal energy here? I'm uncertain of the final volume.

 

[Lagraaaange]

Is there an organized question here, or is it just stream of consciousness which you are writing down?

I have to calculate the change in internal energy of the steam. My main problem is obtaining the final volume of steam.

 

[Chestermiller]

The H is given as the heat of vaporization which is the change in enthalpy. How do I get internal energy here? I'm uncertain of the final volume.

What is the definition of ΔH in terms of ΔU and Δ(PV)?

 

[Lagraaaange]

Lvap = H = U + PV => dH = dU + PdV. I just don't see how the given information will allow me to compute this. I feel that I need the density of steam or specific volume. I don't see what I'm missing here. This is actually an old exam problem from another school so I'm just doing it for practice for my exam tomorrow.

 

[Chestermiller]

Lvap = H = U + PV => dH = dU + PdV. I just don't see how the given information will allow me to compute this. I feel that I need the density of steam or specific volume. I don't see what I'm missing here. This is actually an old exam problem from another school so I'm just doing it for practice for my exam tomorrow.

From the ideal gas law, what is the volume of 1 kg of water vapor at 1 atm and 100 C?
What is the volume of 1 kg of liquid water at 1 atm and 100 C?

 

[Lagraaaange]

density of water = 1000kg/m^3 => V = 1E-3m^3.
vapor: V = nRT/P. n = 1kg/16kg kilomoles.
Then dU = Lm - P(Vvapor - Vwater)?

 

[Chestermiller]

density of water = 1000kg/m^3 => V = 1E-3m^3.
vapor: V = nRT/P. n = 1kg/16kg kilomoles.
Then dU = Lm - P(Vvapor - Vwater)?

Sure. That's right.

Incidentally, I don't understand your answer for the volume of the vapor. It should have units of m^3. I comes out to about 1.67 m^3. Notice that the volume of the liquid water is negligible compared to the volume of the water vapor.

Chet

 

[Lagraaaange]

Sure. That's right.

Incidentally, I don't understand your answer for the volume of the vapor. It should have units of m^3. I comes out to about 1.67 m^3. Notice that the volume of the liquid water is negligible compared to the volume of the water vapor.

Chet

I think I didn't make it too clear. I was just specifying what n would be; not volume. Thanks.

 

[Chestermiller]

I think I didn't make it too clear. I was just specifying what n would be; not volume. Thanks.

n = 1000gm/18 = 55.6 moles