In a closed container at constant pressure, #"N"_2# gas expands from #"49884 cm"^3# to #"62355 cm"^3# when heated to #"395 K"# from an initial temperature. Given that #C_P = "1.014 kJ/kg"cdot"K"#, what is the change in enthalpy for this heating process?

#DeltaH ~~ "2.304 kJ/mol"# added to #"N"_2# gas.

What is the actual amount of heat involved? (#"kJ"#, not #"kJ/mol"#)

Since you know the mols and pressure of the gas are constant, you can start from the ideal gas law to find:

#PV_1 = nRT_1#
#PV_2 = nRT_2#

Dividing these gives Charles's law:

#V_1/T_1 = V_2/T_2#

So, the first temperature, which you need, was:

#T_1 = (V_1/V_2)T_2#

#= ("49884 cm"^3)/("62355 cm"^3) cdot "395 K"#

#=# #"316 K"#

Now, you can use the definition of the change in enthalpy, or the heat flow at constant pressure:

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

#= int_(T_1)^(T_2) C_PdT#

where #C_P# is the constant-pressure heat capacity of the gas. We assume that #C_P# does not vary in the temperature range.

Therefore:

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

This integral easily gives:

#color(blue)(DeltaH) ~~ C_P(T_2 - T_1)#

#= "1.041 kJ"/(cancel"kg"cdotcancel"K") xx cancel"1 kg"/(1000 cancel"g") xx (28.014 cancel"g")/"1 mol N"_2 xx (395 cancel"K" - 316 cancel"K")#

#= color(blue)("2.304 kJ/mol N"_2)#

This means #"2.304 kJ"# of heat went into #"N"_2# to expand it at constant pressure for every #"mol"# of #"N"_2# there was.