Steam is compressed reversibly to liquid water at the boiling point 100C. The heat of vaporisation of water at 100C and 1.01325 bar is 2258Jg^-1. Calculate for one mole of water in term of work done(w),q,H and U for the process
To calculate the work done, heat transferred (q), enthalpy change (ΔH), and internal energy change (ΔU) for the process of steam being compressed reversibly to liquid water at the boiling point of 100°C, we need to use the following equations:
1. Work done (w) = -PΔV
2. Heat transferred (q) = ΔH
3. Enthalpy change (ΔH) = q
4. Internal energy change (ΔU) = q - PΔV
Given:
- Heat of vaporization (ΔHvap) = 2258 J/g
- Boiling point (T) = 100°C
- Pressure (P) = 1.01325 bar
First, we need to convert the heat of vaporization from grams (g) to moles (mol):
- Molar mass of water (H2O) = 18.01528 g/mol
- Heat of vaporization (ΔHvap) in J/mol = ΔHvap in J/g * (1 mol / molar mass of water)
Calculating:
ΔHvap = 2258 J/g * (1 mol / 18.01528 g/mol)
ΔHvap = 125.04 J/mol
Now, let's calculate each quantity:
1. Work done (w):
Since the process is reversible, we assume the external pressure is equal to the given pressure, P = 1.01325 bar.
- Change in volume (ΔV):
At the boiling point, the molar volume of water vapor is approximately 22.4 L/mol.
- ΔV = Vfinal - Vinitial
- ΔV = (liquid state volume) - (vapor state volume)
- ΔV = (1 mol / molar volume of liquid water) - (1 mol / molar volume of water vapor)
Substituting the values:
- ΔV = (1 mol / 18.01528 g/mol * density of liquid water) - (1 mol / 22.4 L/mol)
(Note: Density of liquid water at 100°C is approximately 0.958 g/cm³, which is equivalent to 958 kg/m³.)
Now, let's calculate ΔV in m³:
- ΔV = (1 mol / 18.01528 g/mol * 958 kg/m³) - (1 mol / 22.4 L/mol)
- ΔV = (1 mol / 0.01801528 kg) - (1 mol / 0.0224 m³)
- ΔV = (55.502 mol/kg) - (44.643 mol/m³)
- ΔV = 10.859 mol/m³
Finally, calculate the work done:
w = -PΔV
w = -(1.01325 * 10⁵ Pa) * 10.859 mol/m³
(Note: 1 bar = 10⁵ Pa)
2. Heat transferred (q):
q = ΔHvap
q = 125.04 J/mol
3. Enthalpy change (ΔH):
ΔH = q
ΔH = 125.04 J/mol
4. Internal energy change (ΔU):
ΔU = q - PΔV
Substituting the values:
ΔU = 125.04 J/mol - (-(1.01325 * 10⁵ Pa) * 10.859 mol/m³)
After performing the calculations, you will determine the values for w, q, ΔH, and ΔU in terms of the given units and the molar quantities of water.