The figure shows a cube of air at pressure 1.0 atm above a hot surface on Earth. Contact with the hot surface causes the volume of the air to expand from 2.00 × 105 L to 2.20 × 105 L by the transfer of 8.2 × 106 J of heat. Assume the systems are isolated from the outside world but not from each other.

Question

b. Determine the change in energy for the air. Show your calculations. (Use the conversion 1 L· atm = 101.3 J.)
Delta E = q + w
W = -p * delta V

W = -1 atm * 2.0 * 10^4 = -2.0 * 10^4 L*atm * 101.3 J= 2026000 J
Delta E = 2026000 J *8.2 × 106 J = 1.66 * 10^13 J.
I was told that I made some calculation errors please help me find them.

Answer 1

You have tried it with correct approach. Rest was perfect but the multiplication sign.
Determine the change in energy for the air. Show your calculations. (Use the conversion 1 L· atm = 101.3 J.)
Delta E = q + w
W = -p * delta V

W = -1 atm * 2.0 * 10^4 = -2.0 * 10^4 L*atm * 101.3 J= 2026000 J
Delta E = 2026000 J *8.2 + 106 J = -19894 J
Or -1.98 x 10^4 J

Answer 2

The process is correct but since W is negative it is supposed to be -2026000 J.

Answer 3

And for part be the equation is Delta E = q+w but as w is negative it would be (-). So the proper equation would be 8.2x10^6 - 2026000J.