Thermodynamics: the study of systems involving energy in the form of heat and work.

The First Law of Thermodynamics: Q = ΔE_int + W

Q represents heat added to a system (or removed if it is negative)

ΔE_int is the change in internal energy of the system. This is directly proportional to the change in temperature.

W is the work done by the system.

The First Law is often written as ΔE_int = Q - W

What is temperature?

Temperature is a measure of the internal energy of a system. Internal energy is kinetic energy associated with the motion of atoms, molecules, and electrons making up the system.

What is heat?

Heat is energy transferred between a system and its surroundings because of a temperature difference between them. Heat naturally flows from higher temperature to lower temperature.

What is the connection between them?

The First Law of Thermodynamics links temperature (via internal energy), heat, and work. Adding heat generally corresponds to an increase in temperature, although not always. Examples of exceptions include:

• If all the heat goes into doing work, the temperature will not change.
• During a change of state the internal energy increases without any change in temperature.

We defined work previously as:

W = ∫ F dx

F = PA, so:

W = ∫ PA dx = ∫ P d(Ax) = ∫ P dV

At constant pressure the work done by the system is the pressure multiplied by the change in volume.

If there is no change in volume, no work is done.

We often sketch pressure-volume graphs when we're trying to understand a thermodynamic system. These graphs are helpful because the work done by the system is the area under the P-V graph.

A gas in a cylinder occupies a volume of 0.065 m³ at room temperature (T = 293 K). The gas is confined by a piston with a weight of 100 N and an area of 0.65 m². The pressure above the piston is atmospheric pressure.

(a) What is the pressure of the gas?

This can be determined from a free-body diagram of the piston. The weight of the piston acts down, and the atmosphere exerts a downward force. These two forces are balanced by the upward force coming from the gas pressure. The piston is in equilibrium, so the forces balance. Therefore:

PA = P_atmA + mg

Solving for the pressure of the gas gives:

P = Patm +

mg A

= 101300 +

100 0.65

= 101450 Pa

The pressure in the gas isn't much bigger than atmospheric pressure, just enough to support the weight of the piston.

(b) The gas is heated, expanding it and moving the piston up. If the volume occupied by the gas doubles, how much work has the gas done?

Assume the pressure is constant. Once the gas has expanded and come to a new equilibrium position the pressure will be the same because the free-body diagram is the same. As long as the expansion takes place slowly, it is reasonable to assume that the pressure is constant during the expansion.

At constant pressure the work done is simply:

W = PΔV

W = 101450 * 0.065 = 6590 J

(c) What is the final temperature of the gas?

If the volume doubles while the pressure stays constant, the temperature must also double.