A gas absorbs 10J of heat and is simulataneously comptessed by a constant external pressure of 0.5atm from 4L to2L in volume. What is change in the internal energyA. 101J B. 111J C. 2J D. None of the above

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.