Answer:
a) The work done is 10.0777 kJ
b) The water's change in internal energy is -122.1973 kJ
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:
The work done:
The change in internal energy:
This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C
Nitrogen in the limiting reactant x
Answer:
D
Explanation:
On the left hand side there are a total of 4 hydrogen and 2 oxygen but on the right hand side there Is only 2 hydrogen and 1 oxygen
