What is value of ∆E(heat change at constant volume) for reversible isothermal evaporation of 90g water at 100°C? assuming water vapour behaves as an ideal gas ΔHvapwater =540cals/g.
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a
9×103 cals
b
6×103 cals
c
4.49 cals
d
44870 cals
answer is D.
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Detailed Solution
(ΔH)v=m×Lv=90×540=486000cal ΔH=ΔE+PVV-VL=ΔE+nRT or 48600=ΔE+9018×2×373 or ΔE=48600-3730=44870cals