Work done on steam during phase change?

In summary, the amount of work 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 is -P * (V2 - V1) = -(101325 Pa) * ((.1190 m^3) - (.07002 m^3)) = -4963 J.
  • #1
XianForce
16
0

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 :)
 
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  • #2
XianForce said:
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
(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
It would be very hard to pick up a m3 of water! So, is the density of water 1 kg/m3?
 
  • #5
TSny said:
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
 

1. What is work done on steam during phase change?

Work done on steam during phase change is the amount of energy required to change the state of water from liquid to gas, or vice versa. This process is known as vaporization or condensation, and it involves breaking or forming intermolecular bonds between water molecules.

2. How is work done on steam during phase change calculated?

The work done on steam during phase change can be calculated using the formula W = mL, where W is the work done, m is the mass of the substance, and L is the latent heat of vaporization or condensation. The latent heat is a constant value specific to the substance and can be found in reference tables.

3. What factors affect the work done on steam during phase change?

The work done on steam during phase change is affected by several factors, including the initial temperature of the substance, the amount of energy supplied, and the atmospheric pressure. These factors can impact the amount of energy required to overcome the intermolecular forces and change the state of water.

4. Can the work done on steam during phase change be negative?

Yes, the work done on steam during phase change can be negative. This occurs when the substance is undergoing a change from gas to liquid, and energy is released as the intermolecular bonds are formed. In this case, the work done is considered to be negative because energy is being released rather than added.

5. How does the work done on steam during phase change affect the overall energy of the system?

The work done on steam during phase change is a transfer of energy from or to the system. If energy is added to the system, the work done is positive, and the overall energy of the system increases. Conversely, if energy is released from the system, the work done is negative, and the overall energy of the system decreases.

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