Certainly! Below is the AP Physics 2 Exam question related to the thermodynamic processes:

Question: A sample of an ideal gas is taken through a thermodynamic process. The initial pressure, volume, and temperature of the gas are 2 atm, 4 L, and 300 K, respectively. During the process, 500 J of heat is added to the gas, and the gas expands at constant pressure to a final volume of 8 L. Calculate the change in internal energy (ΔU) of the gas during this process.

A. -250 J B. 250 J C. 500 J D. 750 J

Answer: The change in internal energy (ΔU) for an ideal gas can be calculated using the first law of thermodynamics:

ΔU = Q - W

Where, ΔU = Change in internal energy Q = Heat added to the gas W = Work done by the gas

First, let's calculate the work done by the gas during the process. For a constant pressure process, the work done is given by:

W = PΔV

Where, P = Pressure ΔV = Change in volume

Given that the pressure is constant at 2 atm and the change in volume is ΔV = Vf - Vi = 8 L - 4 L = 4 L, we can calculate the work done:

W = (2 atm)(4 L) = 8 atm*L

Next, we use the ideal gas law to convert atm•L to Joules:

1 atm•L = 101.3 J Therefore, 1 atm•L = 101.3 J

Now, we can convert the work from atm•L to Joules: W = (8 atmL)(101.3 J/atmL) W = 810.4 J

Now, we can calculate the change in internal energy using the first law of thermodynamics: ΔU = Q - W ΔU = 500 J - 810.4 J ΔU = -310.4 J

Rounding to two significant figures, the change in internal energy (ΔU) for the gas during this process is approximately -310 J.

Therefore, the correct answer is not listed. A new answer should be provided and explained.