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Work done on steam during phase change?

  • Thread starter XianForce
  • Start date
  • #1
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Homework Statement


(a) How much work is done on the steam when 3.89 mol of water at 100°C boils and becomes 3.89 mol of steam at 100°C at 1.00 atm pressure? (Assume the latent heat of vaporization of water is 2.26 106 J/kg.)

(b) Assume the steam to behave as an ideal gas. Determine the change in internal energy of the system of the water and steam as the water vaporizes.


Homework Equations



Q + W = ΔE
Q = m * Lv
W = -P * ΔV = -P * (V2 - V1)
P * V = n * R * T

The Attempt at a Solution



Convert moles of water to volume (multiply by molar mass, then use density of water)
Volume of water: 0.07002 m^3.

Then find the volume of steam using ideal gas law: V = ((3.89 mol) * (8.31 J / mol K) * (373.15K)) / (101325 Pa)
Volume of steam: 0.1190 m^3

So work should be -P * (V2 - V1) -> -(101325 Pa) * ((.1190 m^3) - (.07002 m^3))

W = -4963 J = -4.963 kJ.

But my web assign homework tells me that I am over 10% away from the correct answer.

I can't do part (b) until I can complete part (a) because W + Q = ΔE, so without knowing the work, I can't find the change in internal energy (at least that's what my prof said when I asked him).

Thanks in advance for any help :)
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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Convert moles of water to volume (multiply by molar mass, then use density of water)
Volume of water: 0.07002 m^3.)
Check where the decimal point should go here.
 
  • #3
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(3.89 mol) * (18 g / mol) = 70.02 g = 70.02 * 10^(-3) kg

move the decimal over three places: 0.07002 kg, then density of water is 1 kg/ m^3, so:

.07002 m^3, right?
 
  • #4
TSny
Homework Helper
Gold Member
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It would be very hard to pick up a m3 of water! So, is the density of water 1 kg/m3?
 
  • #5
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It would be very hard to pick up a m3 of water! So, is the density of water 1 kg/m3?
Oh my gosh, I can't believe I did that XD! Thanks 1000 kg/m^3
 

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