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.