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 35.0 atm and releases 70.7 kJ of heat. Before the reaction, the volume of the system was 8.80 L . After the reaction, the volume of the system was 2.00 L .

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

Answer 1

q   = -70.7KJ (heat released)

q   = -70700J

p = 35atm

v1 = 8.80L

V2   = 2L

w   =- pΔv

     = -35atm(2-8.8)L

       = 238L-atm   

         = 238*101.3J                           [ 1 L-atm = 101.3J]

          = 24109.4J

ΔE = q + w

        = -70700+24109.4

        = -46590J

          = -46.59KJ

The total internal energy change   = -46.59KJ