Question

The value for the reverible isothermal evaporation of 90gm water at 100C' will be (DELTA H evap. OF WATER =40.8kJ/mol)and R=8.314 J perkelvin per mole​

Answer 1

Answer:

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Answer 2

Answer:

Given,

Water⟶vapour

Moles before evaporation

18

90

=5 0

The evaporation of 5 moles of water takes place reversibly and isothermally into vapours.

Thus, heat given at constant pressure:

ΔH=heatofevaporation×amountevaporated

=540×90

ΔH=48600cal

Also, ΔH=ΔU+Δn

(g)

RT

ΔU=48600−Δn

(g)

RT

=48600−5×2×373

=48600−3730

ΔU=44870cal