Question

1.65 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 m^3⋅mol^−1 and the molar volume of steam at 373 K and 610. K is 3.03 and 5.06 ×10^−2 m^3⋅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. ΔHvap for water is 40.65 kJ⋅mol^−1. Part A) Calculate q for this process. Express your answer with the appropriate units. Part B) Calculate ΔH for this process. Express your answer with the appropriate units.

Answer 1

Answer:

ΔH = q = 8.02 x $10^{4} J$

Explanation:

q = ΔH = nΔH vaporization + nCp,m steam . ΔT

= 1.65 mol x 40656 J $mol^{-1}$ + 1.65 mol x 33.58 J $mol^{-1} K^{-1}$ x (610K - 373K) = 8.02 x $10^{4} J$

w = - P external ΔV = $-10^{5} Pa$ x (1.65 x 5.06 x $10^{-2} m^{3}$ - 1.65 x 1.89 x $10^{-5} m^{3}$) = -8.34 x $10^{3}$ J

ΔU = w + q = -8.34 x $10^{3}$ J + 8.02 x $10^{4} J$ = 7.18 x $10^{4} J$