Question

The vaporization of 1 mole of liquid water (system) at 100.9 C, 1.00 atm, is endothermic.

H2O(L) + 40.7 Kj ----> H2O (g).
Assume at exactly 100.0°c and 1.00 atm total pressure, 1.00 mole of liquid water and 1.00 mole of water vapor occupy 18.80 ml and 30.62 l, respectively.
The calculated work done on or by the system when 3.25 mol of liquid water vaporizes is -10077J.
Calculate the water's change in internal energy.

Answer 1

a) The work done is 10.0777 kJ

b) The water's change in internal energy is -122.1973 kJ

Step-by-step explanation:

Given data:

1 mol of liquid water

T₁ = temperature = 100.9°C

P = pressure = 1 atm

Endothermic reaction

T₂ = temperature = 100°C

1 mol of water vapor

VL = volume of liquid water = 18.8 mL = 0.0188 L

VG = volume of water vapor = 30.62 L

3.25 moles of liquid water vaporizes

Q = heat added to the system = -40.7 kJ

Questions: a) Calculate the work done on or by the system, W = ?

b) Calculate the water's change in internal energy, ΔU = ?

Heat for 3.25 moles:

`Q_(1) =3.25*(-40.7)=-132.275kJ`

The work done:

`W=-nP*delta(V)=-3.25*101.33*(0.0188-30.62)=10077.6637J=10.0777kJ`

The change in internal energy:

`delt(U)=W+Q_(1) =10.0777-132.275=-122.1973kJ`