Careful. There is no "expansion" if the tank is rigid. It's the P part of @Dale 's formula that counts. V would be constant.hjb39 said:I assume the rate of expansion is constant
The relationship between temperature and pressure in a contained volume of air is known as Charles's Law. According to this law, as the temperature of a gas increases, so does its pressure, assuming that the volume and number of particles of the gas remain constant.
Heating air causes the molecules in the air to gain kinetic energy and move faster. This increase in molecular motion leads to more collisions with the walls of the container, resulting in a higher rate of pressure increase within the container.
Some factors that can affect the rate of pressure increase of contained air being heated include the initial temperature and pressure of the air, the volume of the container, and the amount of heat being added to the system.
The rate of pressure increase of air being heated is directly proportional to the temperature increase, while the rate of pressure decrease of air being cooled is directly proportional to the temperature decrease. In other words, heating air will cause its pressure to increase at a faster rate than cooling it will cause its pressure to decrease.
As the volume of the container decreases, the rate of pressure increase of air being heated also increases. This is because the molecules of air have less space to move around, resulting in more frequent collisions with the walls of the container and a higher rate of pressure increase.