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Determine the value of `DeltaH` and `DeltaU` for the reversible isothermal evaporation of `90.0g` of water at `100^(@)C`. Assume that water behaves as an ideal gas and heat of evaporation of water is `540 cal g^(-1) (R = 2.0 cal mol^(-1)K^(-1))`. |
Answer» Total heat change<br> `DeltaH = 90.0 xx 540 = 48600cal`<br> Now, `DeltaH = DeltaU +P DeltaV`<br> Here, `DeltaV =(V_(vapour) -V_(liquid)) = V_(vapour)`<br> (Volume of liquid is negligible as compared to volume of vapor)<br> `DeltaH = DeltaU +PV_(vapour) = DeltaU +nRT`<br> `n = (90)/(18) = 5mol, R = 2 cal K^(-1) mol^(-1), T = 373 K`<br> `:. DeltaH = DeltaU +5 xx2 xx 373`<br> `DeltaH = DeltaU +3730`<br> or `DeltaU = DeltaH - 3730 =48600 - 3730 = 44870cal` | |