Question

a) Find $\Delta$U for the change in state of 1.0 mol H₂O (l) at 83 C to H₂O (g) at 114 C. The molar heat capacity of H₂O (l) = 75.3 J mol^-1K^-1, molar heat capacity of H₂O (g) = 25.0 J mol^-1K^-1, and the heat of vaporization of H₂O is 40.7 x 10³ J mol^-1K^-1 at 100 C.

b) Find ΔH for the change in state of 1.0 mol H₂O(l) at 83 ∘C to H₂O(g) at 114 ∘C.

c) Find q for the change in state of 1.0 molH₂O(l) at 83 ∘C to H₂O(g) at 114 ∘C.

d) Find w for the change in state of 1.0 molH₂O(l) at 83 ∘C to H₂O(g) at 114 ∘C.

Answer 1

To determine the energy involved in converting water at 83ºC into vapor at 114ºC, three stages need consideration. Each stage involves a type enthalpy change (∆H). Ultimately, you’re determining the total amount of energy involved (Q). I believe, the question is asking you to break the problem up into steps. You could use symbols such as ∆U etc, but ∆H is what is most commonly used.

Data:
ΔH (vaporisation) = 40.7 kJ/mol
Cw = Heat capacity water = 75.3 J/m-k = 4.2 J/g-K
Cv = heat capacity of vapor = 25 J/m-k = 2.0 J/g-K

The energy needed to heat the water from 83ºC to 100ºC is calculated using:
∆H = m(Cw)∆T
where:
m = mass of water (g) = 1 mole
Cw = 75.3 J/m-K
∆T = 100 – 83= 17ºC

∆H = m(Cw)∆T = 1 x 75.3 x 17 = 1285.2 J = 1.2852 kJ

The energy needed to vaporise the water is determined using:
H2O(l) → H2O(g) (100°C) ΔH (v) = 40.7 kJ / mole
∆H(v) = 40.7 kJ

The energy needed to heat water from 100ºC to 114ºC requires the same equation as used before, but, using a different heat capacity:
∆H = m(Cv)∆T = 1 x 25 x 14 = 350 J = 0.350 kJ

Hence, the total energy involved is:

q = 1.2852 + 40.7 + 0.350 = 42.33 kJ

∆H = 1.2852 + 0.350 = 1.6352 kJ

Note: Heat capacities do vary with temperature.

w = ∆U - q

with the help of ∆U and q, we can calculate the value of work done (w).

happy learning ?