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?

1 Answer
Oct 15, 2017

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.