Back to chapter

20.2:

Work Done During Volume Change

JoVE Core
Physics
A subscription to JoVE is required to view this content.  Sign in or start your free trial.
JoVE Core Physics
Work Done During Volume Change

Languages

Share

Consider an inflated soccer ball having a volume of V. The pressure acting on the ball is the atmospheric pressure.

Suppose the ball falls off the cliff and gets punctured; the air inside the ball releases. Assuming uniform deflation, let its radius be reduced by a distance of dr.

The magnitude of the force acting on the ball depends on the atmospheric pressure and the surface area of the ball.

Therefore, the work done by the surroundings on the ball equals the negative product of the atmospheric pressure, the ball's surface area, and the change in its radius.

However, the product of surface area and dr is the change in the volume of the ball.

Therefore, during a finite change in volume from V1 to V2, work done can be expressed as the integral of the product of pressure and change in volume.

Here the negative sign indicates that the work is done on the system by the surroundings. Conversely, the work done by the system is considered positive.

20.2:

Work Done During Volume Change

In mechanics, work is done on an object when the force acting on it displaces the object. In thermodynamics, work done on a system can be estimated when the system's volume changes during any thermodynamic process.

Consider a gas confined to a cylinder fitted with a movable piston at one end. If the gas expands from volume V1 to volume V2, it exerts a force on the piston, such that the piston moves by a distance dr.

The work done by the gas on the piston can be expressed as

Equation1

If p is the pressure acting on the face of the piston of area A, then the force acting on the piston is

Equation2

Consequently, the work done can be expressed as

Equation3

The change in the volume of the gas dV equals Adr. Therefore, the work done by the gas on the piston when it expands from volume V1 to V2 is expressed as

Equation4

Here, the integral is only meaningful for a quasi-static process, which means a process that takes place in infinitesimally small steps, keeping the system at thermal equilibrium.

If the piston compresses the gas, the work is done on the gas, and hence conventionally, is considered negative.

Suggested Reading

  1. OpenStax. (2019). University Physics Vol. 1. [Web version]. Pp 112 Retrieved from https://openstax.org/books/university-physics-volume-2/pages/3-2-work-heat-and-internal-energy