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Question

The values of ΔH and ΔU for the reversible isothermal evaporation of 90.0 g of water at 100oC are x cal & y cal. Assume that water vapour behaves as an ideal gas and heat of evaporation of water is 540 cal g1. (R=2cal mol1K1). The value of x+y is:

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Solution

WaterVapour

Mole before evaporation 9018=5

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

Thus, heat given at constant pressure (ΔH) = heat of evaporation x amount evaporated.

ΔH=540×90

ΔH=48600cal

Also, ΔH=ΔU+ΔnRT

ΔU=48600ΔnRT

=48600(5×2×373)

=486003730

ΔU=44870cal

Therefore, ΔH+ΔU=93470

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