The heat of vaporization of water at 100C is 40.66 kj/mol find the change in entropy when 5.0 g of water vapor condenses to liquid at 100 C and 1 atm pressure
-3o.28j/k
s =-30.2j/k
jamal
To find the change in entropy when water vapor condenses to liquid, we need to use the equation:
ΔS = ΔH / T
Where:
ΔS is the change in entropy
ΔH is the heat of vaporization
T is the temperature in Kelvin
First, let's convert the given temperature of 100°C to Kelvin:
T (Kelvin) = 100°C + 273.15 = 373.15 K
Now let's calculate the moles of water vapor using the given mass:
Number of moles = mass (g) / molar mass (g/mol)
The molar mass of water (H2O) is approximately:
Molar mass (H2O) = 2 * (molar mass of hydrogen) + molar mass of oxygen
= 2 * 1.008 g/mol + 15.999 g/mol
= 2.016 g/mol + 15.999 g/mol
= 18.015 g/mol
Number of moles = 5.0 g / 18.015 g/mol
Now that we have the number of moles, we can calculate the change in entropy:
ΔS = ΔH / T
Since the heat of vaporization (ΔH) is given as 40.66 kJ/mol, we need to convert it to J/mol:
ΔH (J/mol) = 40.66 kJ/mol * 1000 J/1 kJ
Finally, we can substitute the values in the equation to find the change in entropy:
ΔS = (ΔH (J/mol)) / T (K)
Calculate the change in entropy using the given values and the formula above, and you will find the answer.