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Water vapor (1.2 kg ) at 0.2 kPa and 30 ∘C is cooled in a constant...

Water vapor (1.2 kg ) at 0.2 kPa and 30 ∘C is cooled in a constant pressure until condensation begins.

a)Water vapor (1.2 kg ) at 0.2 kPa and 30 ∘C is cooled in a constant pressure until condensation begins.

b)Determine the change of enthalpy (ΔH) treating water as the system.

c)What-if scenario: What would the boundary work transfer be if all the vapor condensed?

d)What-if scenario: What would the change of enthalpy be if all the vapor condensed?

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Answer #1

Let us treat the water vapour as polyatomic ideal gas with Specific heat at constant pressure Cp = 4R,

where R is gas constant 8.314 J/ (mole-K)

Number of moles n in 1.2 kg water vapour = 1200/18 = 66.67

(b) change in enthalpy = n Cp\DeltaT = 66.67\times 4 \times 8.314 \times 30 = 66.512 kJ ..................(1)

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(c) we have \Delta Q = \Delta U+ \Delta W

where \Delta Q is heat transfer ( equals to change in enthalpy), \Delta U is change in internal energy

and \Delta W is work done

\Deltau = n Cv\DeltaT = 66.67 \times 3 \times 8.314 \times 30 = 49.886 kJ

Hence, \Delta W = \Delta Q - \Delta U = ( 66.512 - 49.886 ) kJ = 16.63 kJ ...............(2)

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(d) Change in enthalpy as given in eqn.(1) is when the system attains the state at temperature 0oC

and starts condensing by absorbing latent heat.

If all 1.2 kg water is completely condensed, then it has absorbed the latent heat

Hence change in enthalpy = 66.512 + 1.2 \times 2260 = 2779 kJ

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