A gaseous reaction occurs at a constant pressure of 50.0 atm and releases 69.8 kJ of heat. Before the reaction, the volume of the system was 8.80 L . After the reaction, the volume of the system was 2.80 L .
Calculate the total internal energy change, change in U , in kilojoules.
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To calculate the total internal energy change (ΔU) in kilojoules, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
Mathematically, ΔU = q - w
Here, q represents the heat added to the system and w represents the work done by the system.
Given information:
Pressure (P) = 50.0 atm
Heat (q) = -69.8 kJ (negative sign because heat is being released)
Initial volume (V1) = 8.80 L
Final volume (V2) = 2.80 L
To calculate the work done by the system, we can use the formula:
w = -PΔV
where ΔV is the change in volume (V2 - V1). In this case, the system is doing work on the surroundings, so the work will be negative.
Substituting the given values:
ΔV = 2.80 L - 8.80 L = -6.0 L
P = 50.0 atm
w = -50.0 atm * (-6.0 L)
w = 300.0 L.atm
Now, we can calculate the total internal energy change (ΔU):
ΔU = q - w
ΔU = -69.8 kJ - 300.0 L.atm
To convert L.atm to kJ, we use the conversion factor 1 L.atm = 101.3 J:
ΔU = -69.8 kJ - (300.0 L.atm * (101.3 J/1 L.atm))/1000 kJ
Calculating ΔU:
ΔU = -69.8 kJ - 30.39 kJ
ΔU = -100.19 kJ
Therefore, the total internal energy change (ΔU) in this gaseous reaction is -100.19 kJ.