Internal energy and heat of vaporization

[hs764]

1. One gram of water occupies a volume of 1 cm³ at atmospheric pressure. When this amount of water is boiled, it becomes 1671 cm³ of steam. Calculate the change in internal energy for this vaporization process. The heat of vaporization for water is 2.26 x 10⁶ J/kg.

2. w = -PΔV, E = q + w

3. w = -(1 atm)x(1.671 L - 0.001 L) x (101.3 J/L atm) = -169.2 J. q = ΔH_vap = 2.26 x 10⁶ J/kg x 0.001 kg = 2.26 x 10³ J. E = 2.26 x 10³ J - 169.2 J = 2090.8 J.

For some reason I feel like something is missing, but there are no temperature changes given so I don't know what other steps there could be.

 

[ehild]

1. One gram of water occupies a volume of 1 cm³ at atmospheric pressure. When this amount of water is boiled, it becomes 1671 cm³ of steam. Calculate the change in internal energy for this vaporization process. The heat of vaporization for water is 2.26 x 10⁶ J/kg.

2. w = -PΔV, E = q + w

3. w = -(1 atm)x(1.671 L - 0.001 L) x (101.3 J/L atm) = -169.2 J. q = ΔH_vap = 2.26 x 10⁶ J/kg x 0.001 kg = 2.26 x 10³ J. E = 2.26 x 10³ J - 169.2 J = 2090.8 J.

For some reason I feel like something is missing, but there are no temperature changes given so I don't know what other steps there could be.

I do not think there is any problem with your solutions. There are some little things, however. The energy equation is ΔE = q + w. The result should be given by 4 significant digits. Round up that 2090.8 J.