 A rigid tank contains air at 500 kilopascals and 150 degrees Celsius. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65 degrees Celsius and 400 kilopascals respectively. Determine the boundary work during the process. So, we have a rigid tank. And in terms of context clues, a rigid tank implies the volume is constant. And remember that we are describing boundary work. And boundary work is the integral of pressure with respect to volume. If the volume is constant, then our volume differential is zero and this entire term is zero. So the answer to the question is just zero. Now I know what you're thinking, you're thinking, but John, why did we have an entire example problem dedicated to the fact that this is zero? Because if you were encountering this on, say, a stage of a bigger workout analysis, you aren't going to have this so clearly and neatly written out for you. It's not going to be called the boundary work when I so core process. You are going to be evaluating an energy balance on a situation where you have lots of terms appearing. And what you should start to get into the habit of recognizing is that rigid tank means no change in volume. And if we have a process where there is no change in volume, we don't have to consider any boundary work because there is not going to be any boundary work. Don't get caught trying to calculate the boundary work for a process where the volume doesn't move. Because in a process where the boundary doesn't move, there is no work associated with a moving boundary.