for the vaporization of water at 1.00 atm, delta H=43.54 kj/mol at 298 K and delta H= 40.68 kj/mol at 373 K. the constant-pressure heat capacity of liquid water is 75.3 J/molK. calculate the constant pressure heat capacity for H2O(g).

sdas

To calculate the constant pressure heat capacity (Cp) for water vapor (H2O(g)), we can use the equation:

Cp(g) = Cp(l) + ΔHvap / (ΔT)

Where:
Cp(g) is the constant pressure heat capacity of water vapor.
Cp(l) is the constant pressure heat capacity of liquid water.
ΔHvap is the enthalpy of vaporization of water.
ΔT is the difference in temperature (T2 - T1) between the boiling point and the initial temperature.

Given data:
Cp(l) = 75.3 J/molK
ΔHvap at 298 K = 43.54 kJ/mol (Note: Convert it to J/mol by multiplying by 1000)
ΔHvap at 373 K = 40.68 kJ/mol (Note: Convert it to J/mol by multiplying by 1000)

Step 1: Convert ΔHvap values from kJ/mol to J/mol.
ΔHvap at 298 K = 43.54 kJ/mol × 1000 = 43540 J/mol
ΔHvap at 373 K = 40.68 kJ/mol × 1000 = 40680 J/mol

Step 2: Calculate the difference in temperature (ΔT).
ΔT = T2 - T1
ΔT = (373 K) - (298 K)
ΔT = 75 K

Step 3: Plug the values into the equation to calculate Cp(g).
Cp(g) = Cp(l) + ΔHvap / ΔT
Cp(g) = 75.3 J/molK + (43540 J/mol + 40680 J/mol) / 75 K
Cp(g) = 75.3 J/molK + 84220 J/mol / 75 K
Cp(g) = 75.3 J/molK + 1122.93 J/molK
Cp(g) = 76.42 J/molK

Therefore, the constant pressure heat capacity (Cp) for H2O(g) is approximately 76.42 J/molK.