What is value of ∆E(heat change at constant volume) for reversible isothermal evaporation of 90 g water at 100°C. Assuming water vapor behaves as an ideal gas and (∆Hvpn) water - 540 cal g-1
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a
9 × 103 cal
b
6 × 103 cal
c
4.49 cal
d
44870
answer is D.
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Detailed Solution
Heat of system (ΔH)v=m×Lv= 90 × 540 = 486000 calΔH=ΔE+PVV−VL=ΔE+nRT or, 48600=ΔE+9018×2×373 or, ΔE=48600−3730=44870 cals