How much heat at constant volume and at constant pressure is needed to heat #"18 g"# of #"O"_2# from #40^@ "C"# to #100^@ "C"#? #C_V = "12.6 J/mol"cdot"K"#, #C_P = "20.85 J/mol"cdot"K"#.

#"425.28 J"# at constant volume, and #"703.73 J"# at constant pressure.

Enthalpy and internal energy are both functions of temperature.

For any gas whose heat capacities are given, the heat flow involved is given by integrating over the temperature range for:

• the change in enthalpy #H# when at constant pressure (e.g. a coffee-cup calorimeter).

#DeltaH = int_(T_1)^(T_2) ((delH)/(delT))_PdT#

• change in internal energy #U# when at constant volume (e.g. a bomb calorimeter).

#DeltaU = int_(T_1)^(T_2) ((delU)/(delT))_VdT#

By definition, the constant-volume heat capacity #C_V# and the constant-pressure heat capacity #C_P# are #((delU)/(delT))_V# and #((delH)/(delT))_P#, respectively.

We assume they are constant in the temperature range.

That allows us to find the heat flow at constant volume, #q_V#:

#ul(q_V) = DeltaU = int_(T_1)^(T_2) C_VdT#

#~~ C_Vint_(T_1)^(T_2) dT#

#= ul(C_VDeltaT)#

(You should recognize this from General Chemistry.)

This gives:

#color(blue)(q_V) = 18 cancel("g O"_2) xx cancel("1 mol O"_2)/(31.998 cancel("g O"_2)) xx "12.6 J"/(cancel("mol O"_2) cdot cancel"K") xx (373.15 cancel"K" - 313.15 cancel"K")#

#=# #color(blue)ul"425.28 J"#

though you only have two sig figs...

And similarly, the heat flow at constant pressure, #q_P#:

#ul(q_P) = DeltaH = int_(T_1)^(T_2) C_PdT#

#~~ C_P int_(T_1)^(T_2) dT#

#= ul(C_PDeltaT)#

(You should recognize this from General Chemistry.)

Therefore:

#color(blue)(q_P) = 18 cancel("g O"_2) xx cancel("1 mol O"_2)/(31.998 cancel("g O"_2)) xx "20.85 J"/(cancel("mol O"_2) cdot cancel"K") xx (373.15 cancel"K" - 313.15 cancel"K")#

#=# #color(blue)ul"703.73 J"#

In this case, there is a significant difference in #DeltaH# vs. #DeltaU# due to the #Delta(PV)# term in the heating process. We then find that we have:

#overbrace(DeltaH)^("703.73 J") = overbrace(DeltaU)^("425.28 J") + overbrace(PDeltaV + VDeltaP)^("278.45 J")#

So in principle, if we knew the starting pressure and volume during the constant-volume heating, and the final pressure afterwards, we could find the final volume after the constant-pressure heating.