Question

27. Determine the value of H and U for reversible isothermal evaporation of 90.0g of water

at 1000C. Assume that water evaporation of water is 540 Cal (K=2.0 Cal mol-1

k

-1

)​

Answer 1

Answer:

Explanation:

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