Question

Some heat is added to a gas container that is topped by a movable piston. The piston is weighed down with a 2 kg mass. The piston rises a distance of 0.2 m at a constant velocity. Throughout this process, the temperature of the gas in the container remains constant. How much heat was added to the container?

Answer 1

The key to answering this question is to note that the temperature of the container remains constant. That means that the internal energy of the system remains constant$(\Delta U=0)$, which means that, according to the First Law,$\Delta Q+\Delta W=0$.By pushing the piston upward, the system does a certain amount of work,$\Delta W$and this work must be equal to the amount of heat added to the system,$\Delta Q$.
The amount of work done by the system on the piston is the product of the force exerted on the piston and the distance the piston is moved. Since the piston moves at a constant velocity, we know that the net force acting on the piston is zero, and so the force the expanding gas exerts to push the piston upward must be equal and opposite to the force of gravity pushing the piston downward. If the piston is weighed down by a two-kilogram mass, we know that the force of gravity is:

$F=mg=2\times 9.8=19.6N$

1Since the gas exerts a force that is equal and opposite to the force of gravity, we know that it exerts a force of 19.6 N upward. The piston travels a distance of 0.2 m, so the total work done on the piston is: Since ΔWΔW in the equation for the First Law of Thermodynamics is negative when work is done on the system and positive when work is done by the system, the value of ΔWΔW is 3.92 J. Because ΔU=0ΔU=0, we can conclude that ΔQ=3.92ΔQ=3.92J, so 3.92 J of heat must have been added to the system to make the piston rise as it did.