Dispersal of Matter and Energy

As we extend our discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas on the left and a vacuum on the right (see the figure below). When the valve is opened, the gas spontaneously expands to fill both flasks. Recalling the definition of pressure-volume work from the tutorial on thermochemistry, note that no work has been done because the pressure in a vacuum is zero.

\(w=\text{−}P\text{Δ}V=0\phantom{\rule{5em}{0ex}}(P=\text{0 in a vaccum})\)

Note as well that since the system is isolated, no heat has been exchanged with the surroundings (q = 0). The first law of thermodynamics confirms that there has been no change in the system’s internal energy as a result of this process.

\(\text{Δ}U=q+w=0+0=0\)

The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the driving force appears to be related to the greater, more uniform dispersal of matter that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous process took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask).

A diagram shows two two-sided flasks connected by a right-facing arrow labeled “Spontaneous” and a left-facing arrow labeled “Nonspontaneous.” Each pair of flasks are connected to one another by a tube with a stopcock. In the left pair of flasks, the left flask contains thirty particles evenly dispersed while the right flask contains nothing and the stopcock is closed. The right pair of flasks has an open stopcock and equal numbers of particles in both flasks.

An isolated system consists of an ideal gas in one flask that is connected by a closed valve to a second flask containing a vacuum. Once the valve is opened, the gas spontaneously becomes evenly distributed between the flasks.

Now consider two objects at different temperatures: object X at temperature T_X and object Y at temperature T_Y, with T_X > T_Y (see the figure below). When these objects come into contact, heat spontaneously flows from the hotter object (X) to the colder one (Y). This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y.

\({q}_{\text{X}}<0\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{q}_{\text{Y}}=\text{−}{q}_{\text{X}}>0\)

From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a more uniform dispersal of energy.

Two diagrams are shown. The left diagram is comprised of two separated squares; the left is red and labeled “X” and the right is blue and labeled “Y.” Below this diagram is the label “T subscript X, a greater than sign, T subscript Y.” The right diagram shows the boxes next to one another, shaded red on the left, blue on the right, and blended red and blue together in the middle. The left box is red and labeled “X,” the right is blue and labeled “Y” and a right-facing arrow labeled “Heat” is written above them. Below this diagram is the label “X and Y in contact.

When two objects at different temperatures come in contact, heat spontaneously flows from the hotter to the colder object.

As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy.

Example

Redistribution of Matter during a Spontaneous Process

Describe how matter is redistributed when the following spontaneous processes take place:

(a) A solid sublimes.

(b) A gas condenses.

Solution

This figure has three photos labeled, “a,” “b,” and “c.” Photo a shows a glass with a solid in water. There is steam or smoke coming from the top of the glass. Photo b shows the bottom half of a glass with water sticking to its outside surface. Photo c shows a sealed container that holds a red liquid.

(credit a: modification of work by Jenny Downing; credit b: modification of work by “Fuzzy Gerdes”/Flickr; credit c: modification of work by Sahar Atwa)

(a) Sublimation is the conversion of a solid (relatively high density) to a gas (much lesser density). This process yields a much greater dispersal of matter, since the molecules will occupy a much greater volume after the solid-to-gas transition.

(b) Condensation is the conversion of a gas (relatively low density) to a liquid (much greater density). This process yields a much lesser dispersal of matter, since the molecules will occupy a much lesser volume after the gas-to-liquid transition.

(c) The process in question is dilution. The food dye molecules initially occupy a much smaller volume (the drop of dye solution) than they occupy once the process is complete (in the full glass of water). The process therefore entails a greater dispersal of matter. The process may also yield a more uniform dispersal of matter, since the initial state of the system involves two regions of different dye concentrations (high in the drop, zero in the water), and the final state of the system contains a single dye concentration throughout.

Dispersal of Matter and Energy