Question

An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 40.0 atm and releases 67.4 kJ of heat. Before the reaction, the volume of the system was 6.40 L . After the reaction, the volume of the system was 3.00 L .

Calculate the total internal energy change, ΔE, in kilojoules.

Answer 1

First of all, lets calculate the work. This is a constant pressure process. Thus the work done is equal to;

W= - PV

W= - (40x101325 Pa x (3-6.4)x10^-3 m³)

W= 13.78 kJ

As this is a volume contraction, the work is done on the system; hence the positive vale.

Finally, we can use the 1st law of thermodynamics to calculate the internal energy change. According to the frst law of thermodynamics,

E=Q+W

where E is the internal energy, Q is heat and W is work.

E= - 67.4kJ + 13.78kJ

E= - 53.62

Note the negative sign for heat. This is because heat is released from the system.