What is delta U when 2.0mole of liquid water vapourises at 100 degree celcius? The heat of vapourisation, Delta H vap. Of water at 100 degree celcius is 40.66 KJ/Mol
To find the change in internal energy (ΔU) when 2.0 moles of liquid water vaporize at 100 degrees Celsius, we need to use the equation:
ΔU = n * ΔH
Where:
ΔU is the change in internal energy
n is the number of moles of the substance
ΔH is the enthalpy change of the substance (in this case, the heat of vaporization)
We are given:
n = 2.0 moles
ΔH = 40.66 kJ/mol
Substituting these values into the equation, we get:
ΔU = 2.0 moles * 40.66 kJ/mol
To find the value of ΔU, multiply:
ΔU = 81.32 kJ
Therefore, ΔU when 2.0 moles of liquid water vaporize at 100 degrees Celsius is 81.32 kJ.