AP Physics 2 Exam Question

A gas undergoes a thermodynamic process represented by the following PV diagram:

The gas starts at point A and proceeds to point B, where its volume increases while pressure decreases. It then follows a constant pressure process from point B to point C, where the volume decreases. Finally, the gas undergoes an isothermal process from point C to point A.

(a) Calculate the work done by the gas as it undergoes the process from point A to point C. (b) Determine the change in internal energy of the gas as it undergoes the entire process from point A to point C. (c) Calculate the heat absorbed or released by the gas during the entire process.

Answer

(a) To calculate the work done by the gas, we need to determine the area under the curve on the PV diagram. In this case, the process from A to C is not a straight line, so we need to divide it into two parts: AB and BC.

First, consider the process from A to B. The work done during this process is given by the area under the curve AB:

The area of triangle AB is given by:

Next, consider the constant pressure process from B to C. The work done during this process is given by the area under the curve BC, which is a rectangle:

The area of rectangle BC is given by:

The total work done by the gas is the sum of the work done during the two processes:

where $\Delta U$ is the change in internal energy, Q is the heat added or released by the system, and W is the work done on or by the system.

Since the process is cyclic, the change in internal energy is zero ($\Delta U = 0$).

Therefore, the change in internal energy of the gas as it undergoes the entire process from A to C is zero.

(c) To calculate the heat absorbed or released by the gas during the entire process, we can use the equation:

Given that $\Delta U = 0$ (as calculated in part (b)), the heat absorbed or released by the gas is equal to the negative of the total work done:

Substituting the expression for the total work calculated in part (a), we get: