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Determine the value of ΔH and ΔU for the reversible isothermal evaporation of 90.0 g of water at 100° C. Assume that water vapour behave as an ideal gas and heat of evaporation of water is 540 cal.

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â??H = 90.0 x 540 =48600 calVolume of liquid is negligible as compared to volume of vapour.â??H = â??U + ${ pV }_{ vap }$â??H = â??U + nRT= 48600 - 90/18 x 2 x 373(n = 90/18)= 48600 - 3730 = 44870 cal
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 â??H = 90.0 x 540 =48600 cal

Volume of liquid is negligible as compared to volume of vapour.
â??H = â??U + ${ pV }_{ vap }$
â??H = â??U + nRT
= 48600 - 90/18 x 2 x 373(n = 90/18)
= 48600 - 3730 = 44870 cal 

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The total heat change, ΔH=90×540=48600calΔH=90×540=48600cal Now, ΔH=ΔE+PΔVΔH=ΔE+PΔV ΔV=(Vvapor–Vliquid)=Vvapor(Vliquidisnegligible)ΔV=(Vvapor–Vliquid)=Vvapor(Vliquidisnegligible) ΔH=ΔE+PVvapor=ΔE+nRTΔH=ΔE+PVvapor=ΔE+nRT ⇒ΔE=ΔH–nRT⇒ΔE=ΔH–nRT =48600–90/18×2×373=48600–3730=44870cal ...
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The total heat change, ΔH=90×540=48600calΔH=90×540=48600cal
Now, ΔH=ΔE+PΔVΔH=ΔE+PΔV
ΔV=(Vvapor–Vliquid)=Vvapor(Vliquidisnegligible)ΔV=(Vvapor–Vliquid)=Vvapor(Vliquidisnegligible)
ΔH=ΔE+PVvapor=ΔE+nRTΔH=ΔE+PVvapor=ΔE+nRT
⇒ΔE=ΔH–nRT⇒ΔE=ΔH–nRT
=48600–90/18×2×373=48600–3730=44870cal
 
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â??H = 90.0 x 540 =48600 cal â??H = â??U + nRT (R = gas constant,use its value in calorie)= 48600 - 90/18 x 2 x 373(n = 90/18)= 48600 - 3730 = 44870 cal
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