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