# At #100^@ "C"# and #"1.00 atm"#, the molar volume of water vapor is #"30.6 L/mol"# and the enthalpy of vaporization is #"40.66 kJ/mol"#. What is the change in internal energy at this temperature for #"1 mol"# of water? (#rho("H"_2"O"(l)) ~~ "996 g/L"#)

##### 1 Answer

I got

You were given (in order), at

#V_(H_2O(g))/n = "30.6 L/mol"# #(DeltaH_"cond")/(n_(H_2O)) = -q_"vap"/n_(H_2O) = -"40.66 kJ/mol"# , since condensation requires the**release**of heat in order to turn a*less*-ordered phase (gas) into a*more*-ordered phase (liquid).#rho_(H_2O(l)) = "0.996 g/cm"^3 = "0.996 g/mL" = "996 g/L"# is the density of water at the relevant#T# and#P# .

where

You should have the following equation available for a **constant pressure** process:

#bb(DeltaH = DeltaE + PDeltaV)#

Using the definition of *constant pressure*:

#q_P = DeltaE + PDeltaV#

#DeltaE = q_P - PDeltaV#

which you should recognize as the first law of thermodynamics,

#bb(DeltaE_"cond" = q_(P,"cond") - PDeltaV_(H_2O(g->l)))# where

#DeltaV_(H_2O(g->l))# is thechange in volume that water experienceswhen condensing. That is, the initial volume is of theliquid, and the final volume is of thegas. Recall that#q_(P,"cond")# is the heat flow at a constant pressure for a condensation process.

Now, with

#V_(H_2O(l))/n_(H_2O(l)) = M_(m,H_2O(l))/[rho_(H_2O(l))]#

#V_(H_2O(l)) = (n_(H_2O(l))M_(m,H_2O(l)))/[rho_(H_2O(l))]#

#= cancel("1 mol") "water"xx(18.015 cancel"g")/cancel"mol"xx "1 L"/(996 cancel"g") = "0.01809 L"#

You were also given the molar volume of the water vapor, so you can now calculate the **change in volume**:

#color(green)(DeltaV_(H_2O(g->l))) = V_(H_2O(l)) - V_(H_2O(g))#

#= "0.01809 L" - "30.6 L"#

#= -"30.582 L" ~~ color(green)(-"30.6 L")#

which makes sense because **the volume of 1 mol of gas is much larger than the volume of 1 mol of the corresponding liquid**. So, the change in volume is negative, i.e. the water was

*greatly compressed*during condensation.

Note that in general, liquids are amorphous but hard to compress, while gases are amorphous and easy to compress. Thus, you can approximate

You can calculate **conversion factor** constructed by the universal gas constants

#("8.314472 J")/("0.082057 L"cdot"atm") ~~ "101.33 J/L"cdot"atm"#

Therefore:

#color(blue)(DeltaE) = -"40.66 kJ" - [(1.00 cancel"atm")(-30.582 cancel("L")) xx (101.33 cancel"J")/(cancel("L"cdot"atm")) xx "1 kJ"/(1000 cancel"J")]#

#= color(blue)(-"43.8 kJ")#

and due to the number of sig figs, you'd get the same answer even with the assumption of ignoring the molar volume of liquid water.

Or, if you wanted it in