Question

Calculate w, q, ΔH, and ΔU for the process in which 1.78 moles of water undergoes the transition H2O(l,373K)→H2O(g,610.K) at 1 bar of pressure. The volume of liquid water at 373 K is 1.89×10−5 m3 mol−1 and the molar volume of steam at 373 K and 610. K is 3.03 and 5.06 ×10−2 m3 mol−1, respectively. For steam, CP,m can be considered constant over the temperature interval of interest at 33.58 J mol−1 K−1. Assume that the molar enthalpy of vaporization of water is 40656 J mol−1.

Answer 1

The volume of liquid water = 1.89×10−5 m3 mol−1 = 1.78 X 1.89 X 10^-5 m^3 = 3.36 X 10^-5

Volume of vapour at 373 K = 3.03 ×10−2 m3 mol−1 = 1.78 X 3.02 ×10−2 m3 = 5.37 X 10^-2

Volume of vapur at 621K = 5.06 ×10−2 m3 mol−1 = 1.78 X5.06 ×10−2 m3 = 9.01 X 10^-2

so work done = Pressure X change in volume = 1 bar X (9.01 X 10^-2 - 3.36 X 10^-5 ) = 9006.64 X 10^-5 bar m^3

1bar m^3 = 10^5 joules

so 9006.64 X 10^-5 bar m^3 = 9006.64 Joules

Heat = The heat will be required to convert liquid water to vapours + heating up the water vapours

Q1 = Moles of water X molar heat of vapourization = 1.78 X 40656 = 72367.68 J

Q2 = Moles of water X specfic heat of steam X change in temperature = 1.78 X 33.58 X (610-373) = 14166.05 Joules

so total heat = 86533.73 Joules

Change in internal energy = Heat - work = 86533.73 Joules - 9006.64 Joules = 77533.09 Joules

Also the process is accomplished at constant pressure

So heat change at constant pressure = Change in enthalpy = 86533.73 Joules