What are values of ΔH, ΔU, ΔS for 1 mole of H20?
Find ΔH, ΔU, ΔS for 1 mole of H20(l) on T=293K, p=101 325Pa and Cp(l)=75,36 J/molK and for 1 mole of H20(g) when T=520K, p=101 325 Pa, Cp(g)=36kJ/molK and
what is the change in enthalpy of vaporization of water on T=373K and p=101 325Pa and ΔH=40695,7 J/mol
NOTE: #C_P(g) = 36# #J"/"molcdotK# . Seems to be a typo.
- Truong-Son
Find ΔH, ΔU, ΔS for 1 mole of H20(l) on T=293K, p=101 325Pa and Cp(l)=75,36 J/molK and for 1 mole of H20(g) when T=520K, p=101 325 Pa, Cp(g)=36kJ/molK and
what is the change in enthalpy of vaporization of water on T=373K and p=101 325Pa and ΔH=40695,7 J/mol
NOTE:
- Truong-Son
1 Answer
The way the question is worded is quite confusing, but if that's the way it's worded, here's how I'm interpreting this.
Since the definitions of the three thermodynamic functions given are:
#DeltaH_(H_2O(l)) = int_(T_1)^(T_2) C_P(l)dT#
#DeltaS_(H_2O(l)) = int_(T_1)^(T_2) (C_P(l))/TdT#
#DeltaU_(H_2O(l)) = DeltaH_(H_2O(l)) - Delta(PV)#
we can work from there.
It seems reasonable (though tedious) that the question wants us to calculate for
This would illustrate the definitions given above.
CHANGE IN ENTHALPY OF HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR
#DeltaH_(H_2O(l)) = int_("293 K")^("373 K") C_P(l)dT#
In this temperature range, we may assume that
#color(blue)(DeltaH_(H_2O(l))^(293 -> "373 K")) ~~ C_P(l)int_("293 K")^("373 K") dT#
#= ("75.36 J/mol"cdot"K")("373 - 293 K")#
#=# #color(blue)("6028.8 J/mol")#
Include the
Next:
#color(blue)(DeltaH_(H_2O(g))^(373 -> "520 K")) ~~ C_P(g)int_("373 K")^("520 K") dT#
#= ("36 J/mol"cdot"K")("520 - 373 K")#
(not#36000# ; the kilo was a typo.)
#=# #color(blue)("5292 J/mol")#
This gives a total
#bbcolor(blue)(nDeltaH) = n(DeltaH_(H_2O(l))^(293 -> "373 K") + DeltaH_"vap" + DeltaH_(H_2O(g))^(373 -> "520 K"))#
#= ("1 mol")(6028.8 + 40695.7 + 5292) = "52016.5 J"#
#=# #bbcolor(blue)("52.017 kJ")#
CHANGE IN ENTROPY OF HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR
Similarly, the change in entropy over a temperature range can be calculated by the same kind of approximation about
#color(blue)(DeltaS_(H_2O(l))^(293 -> "373 K")) ~~ C_P(l)int_("293 K")^("373 K") 1/TdT#
#= ("75.36 J/mol"cdot"K")ln(373/293)#
#=# #color(blue)("18.19 J/mol"cdot"K")#
At constant temperature and pressure, which is how it is at the phase equilibrium we establish for vaporization,
#color(blue)(DeltaS_"vap") = (DeltaH_"vap")/T_b = ("40695.7 J/mol")/("373 K")#
#=# #color(blue)("109.10 J/mol"cdot"K"#
Finally, a similar process for
#color(blue)(DeltaS_(H_2O(g))^(373 -> "520 K")) ~~ C_P(g)int_("373 K")^("520 K") 1/TdT#
#= ("36 J/mol"cdot"K")ln(520/373)#
#=# #color(blue)("11.96 J/mol"cdot"K")#
So, the total
#bbcolor(blue)(nDeltaS) = n(DeltaS_(H_2O(l))^(293 -> "373 K") + DeltaS_"vap" + DeltaS_(H_2O(g))^(373 -> "520 K"))#
#= ("1 mol")(18.19 + 109.10 + 11.96)# #"J/mol"cdot"K"#
#=# #bbcolor(blue)("139.25 J/K")#
CHANGE IN INTERNAL ENERGY FOR HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR
Recall that
#DeltaH = DeltaU + Delta(PV)#
#= DeltaU + PDeltaV + VDeltaP + DeltaPDeltaV# .
Well, we've been at constant pressure, so we don't have to worry about the
For the heating process, we'd need the densities of water at both temperatures to get accurate molar volumes (which are necessary because liquids are fairly incompressible). I think we can assume that
But just to see...
#rho_("293 K") = "998.21 g/L"#
#rho_("373 K") = "958.37 g/L"#
#barV_("293 K") = M_m/rho = "g"/"mol" xx "L"/"g" = 18.015/998.21 = "0.018047 L/mol"#
#barV_("373 K") = M_m/rho = 18.015/958.37 = "0.018798 L/mol"#
So for
#DeltaH = DeltaU + Delta(PV) = DeltaU + PDeltaV# ,
(at constant pressure)
#color(blue)(nDeltaU_(H_2O(l))^(293 -> "373 K")) = nDeltaH_(H_2O(l))^(293 -> "373 K") - PDeltaV_(H_2O(l))^(293 -> "373 K")#
#=# #("1 mol")("6028.8 J/mol") - ("1.01325 bar")("0.000751 L")xx("8.314472 J")/("0.0831345 L"cdot"bar")#
#=# #color(blue)("6028.7 J") ~~ DeltaH_(H_2O(l))^(293 -> "373 K")#
Indeed, the
For the vaporization, a similar idea follows in that
#PDeltaV_((l)->(g)) ~~ PV_((g)) = nRT#
#= ("1 mol")("8.314472 J/mol"cdot"K")("373 K") = "3101.30 J"#
Therefore:
#color(blue)(nDeltaU_"vap") = nDeltaH_"vap" - PDeltaV_((l)->(g))#
#= ("1 mol")("40695.7 J/mol") - ("3101.30 J")#
#=# #color(blue)("37594.4 J")#
Finally, for the heating of the gas, again, we can assume ideality to get the change in volume of the water vapor. At constant
#PDeltaV = nRDeltaT#
#= ("1 mol")("8.314472 J/mol"cdot"K")("520 - 373 K")#
#=# #"1222.23 J"#
So:
#color(blue)(nDeltaU_(H_2O(g))^(373 -> "520 K")) = nDeltaH_(H_2O(g))^(373 -> "520 K") - PDeltaV#
#= ("1 mol")("5292 J/mol") - ("1222.23 J")#
#=# #color(blue)("4069.77 J")#
So, the overall
#bbcolor(blue)(nDeltaU) = n(DeltaU_(H_2O(l))^(293 -> "373 K") + DeltaU_"vap" + DeltaU_(H_2O(g))^(373 -> "520 K"))#
#= ("1 mol")(6028.7 + 37594.4 + 4069.77)# #"J/mol"#
#=# #bbcolor(blue)("47.693 kJ")#