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 50.0 atm and releases 70.7 kJ of heat. Before the reaction, the volume of the system was 9.00 L . After the reaction, the volume of the system was 2.60 L . Calculate the total internal energy change, ΔE, in kilojoules.

Answer 1

Answer – Given, Pressure = 50.0 atm, V1 = 9.00 L , V2 = 2.60 L ,

Heat, q = -70.7 kJ = -70700 J

First we need to calculate the work

We know, w = =-P*∆V

                         = - 50.0 atm * (2.60-9.0) L

                         = 320 atm.L

We know

1 L.atm = 101.32 J

So, 320 L.atm = ?

= 32422.4 J

So, ∆E = q + w

            = -70700 J + 32422.4 J

            = -38278 J

             = -38.28 kJ