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Question

For the reversible isothermal evaporation of 90.0 g of water at 100C. The value of ΔU for the reaction is:

[Assuming that water vapours behave as an ideal gas and heat of evaporation of water is 540calg1 and R=2calmol1K1]

A
45870cal
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B
56870cal
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C
42000cal
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D
44870cal
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Solution

The correct option is D 44870cal
Given,
Watervapour

Moles before evaporation 9018=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=48600Δn(g)RT

=486005×2×373

=486003730

ΔU=44870cal

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