Question

90 g of water spilled out from a vessel on floor . Asuming water vapour behaving ideal gas. Calculate the internal energy change when the spilled water undergoes complete evaporation at 100celcius .if molar enthalpy of vaporisation of water at 1 bar and 373 k is 41 kj/mol

Answer 1

1 mole water = 18 g

Thus, 90 g = (90/18) = 5 moles

The required formula is,

ΔH = ΔU + ΔnRT

Given,molar enthalpy of vaporisation is  41 kJ/mol [at 1 bar and 373 K]

Thus, for 5 moles, net change in enthalpy = (5 x 41) = 205 kJ/mol

Therefore,

205 x 10³ = ΔU + 5 x 8.314 x 373

or, ΔU = 205000 - 15505.61

or, ΔU = 189494.39

Thus the internal energy change is 189.5 kJ/mole.

Answer 2

Explanation:

90 gram of water is given

so the number of moles is equal to given weight by molar mass

no.of moles =90g/18g=5mol

5H2O (l) gives 5H2O(g)