Principle of Minimum Energy - An Example

An Example

Suppose we have a cylinder containing an ideal gas, with cross sectional area A and a variable height x. Suppose that a weight of mass m has been placed on top of the cylinder. It presses down on the top of the cylinder with a force of mg where g is the acceleration due to gravity.

Suppose that x is smaller than its equilibrium value. The upward force of the gas is greater than the downward force of the weight, and if allowed to freely move, the gas in the cylinder would push the weight upward rapidly, and there would be frictional forces that would convert the energy to heat. If we specify that an external agent presses down on the weight so as to very slowly (reversibly) allow the weight to move upward to its equilibrium position, then there will be no heat generated and the entropy of the system will remain constant while energy is transferred as work to the external agent. The total energy of the system at any value of x is given by the internal energy of the gas plus the potential energy of the weight:

where T is temperature, S is entropy, P is pressure, μ is the chemical potential, N is the number of particles in the gas, and the volume has been written as V=Ax. Since the system is closed, the particle number N is constant and a small change in the energy of the system would be given by:

Since the entropy is constant, we may say that dS=0 at equilibrium and by the principle of minimum energy, we may say that dU=0 at equilibrium, yielding the equilibrium condition:

which simply states that the upward gas pressure force (PA) on the upper face of the cylinder is equal to the downward force of the mass due to gravitation (mg).

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