Answer:
1) ΔE = -800 J, 2) ΔE = 200 J, 3) ΔE = -200 J, 4) ΔE = 800 J
Explanation:
For this exercise let's use the first law of thermodynamics
ΔE = Q + W
Where we will apply this expression to several cases
1) output Q = 500 J and does work of W = 300 J.
The two processes involve an energy output
ΔE = - 500 - 300
ΔE = -800 J
Therefore, the internal energy of the gas decreases by this amount,
2) enter Q = 500J and do work of W = 300 J
Positive heat is added and does negative work
ΔE = 500 - 300
ΔE = 200 J
3) they leave Q = 500 J and do (work on the gas) W = 300J
negative heat and positive work come out
ΔE = -500 + 300
ΔE = -200 J
4) heat enters Q = 500 J, work on gas W = 300 J
both positive
ΔE = 500 + 300
ΔE = 800 J
