Answer:
B. 111 J
Explanation:
The change in internal energy is the sum of the heat absorbed and the work done on the system:
ΔU = Q + W
At constant pressure, work is:
W = P ΔV
Given:
P = 0.5 atm = 50662.5 Pa
ΔV = 4 L − 2L = 2 L = 0.002 m³
Plugging in:
W = (50662.5 Pa) (0.002 m³)
W = 101.325 J
Therefore:
ΔU = 10 J + 101.325 J
ΔU = 111.325 J
Rounded to three significant figures, the change in internal energy is 111 J.
The answers that fit the blanks are SMALL and LITTLE, respectively. The particles or molecules or fas are small which makes it loose and easily moves around, and these only exert little attraction for other gas particles. The answer for this would be option D.
Your answer is probably
Vaporization point
Question 1 answer: A
Question 2 answer: H
Question 3 answer: J
Question 4 answer: T
