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
DeltaH ~~ "2.304 kJ/mol" added to"N"_2 gas.
What is the actual amount of heat involved? (
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 thatC_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