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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Thermodynamics > Path Functions > Work

Work, in terms of thermodynamics, is defined as a movement against a force.

\(W=F \times d\)

In this equation, work is equal to the force applied on the object multiplied by the distance it is displaced. Work is given in joules [J] when the force is in Newtons [N] and the distance is in meters [m]. One joule is the amount of work required to displace an object one meter against a force of one Newton.

The SI units for energy is the joule (J). 1 joule is defined as:

\((Kg*m^2)/(s^2)\)

Calories can also be used as a unit in the answer for work although they are not SI units (: 1 calorie = 4.184 J).

In the above image, the arrow is doing work on the red box. This work is equal to the force applied by the arrow multiplied by the distance the red box is moved.

This is used to measure the relationship between how much energy a material can absorb vs. how much the temperature of the material will change. The amount of work done on a system is equal to the change in the energy of the system.

Measurements of Specific heat Capacity: Energy/(Mass*Temperature)

ie. calorie/(gram*degree)or Joules/(gram*degree)

Therefore, multiplying a specific heat capacity times the mass and the change in temperature of a system, the result will be either the energy gained or lost by the system.

A specific type of work is known as Pressure-Volume work. This occurs when gases are involved. In this special situation a new formula is utilized:

\(W=-p \Delta V\)

In this new equation work is equal to the opposite of the external pressure multiplied by the change in volume of the gas. This change in volume is caused by heat exchange in the system. This value can be found by subtracting the initial volume of the gas from the final volume of the gas after the heat is exchanged. This equation can also be used to determine the energy change in a system as long as the amount of heat exchanged is known. To determine this we must utilize the following equation:

\(\Delta U=q- p\Delta V\)

This concept of volume work is applied in car engines. As gas combusts, heat is released. This is because the reaction is an exothermic reaction. As this heat is released, it does work on the pistons in the engine. This is caused by the gas expanding due to the heat from the reaction. This expansion of gas causes the pistons in the engine to move up and down.

Work can also be done by gasses when external pressure is changed. This change in pressure causes the gas to expand (when pressure is decreased) or to contract (when pressure is increased). This change in pressure causes a change in volume as stated in Boyle's law.

Pressure Volume work is said to be a path function. This means that the system's state is influenced by the path taken to achieve it. In this kind of system the final result is affected by the intermediate steps taken to achieve it. The more expansion steps, the more work can the gas do. When there is an infinite number of infinitely small expansion steps, the reaction is said to be reversible.

- If John pushes a tricycle 50 feet with a force of 5 Newtons how much work did he do on the tricycle?
- How far, in meters, would a box travel if pushed with 7 newtons of force and 15 Joules of work is done on the box?
- How much work is done by a gas that expands from 2 liters to 5 liters against an external pressure of 750 mmHg?
- How much work is done by .54 moles of a gas that has an initial volume of 8 liters and expands under the following conditions: 30
^{o}C and 1.3 atm? - How much work is done by a gas (p=1.7 atm, V=1.56 L) that expands against an external pressure of 1.8 atm?

1. Convert 50 feet to meters: 50 feet x .3048 meters/feet = 15.24 meters.

Plug in equation \W = F \times D\): 5 Newtons * 15.24 meters = 76.2 Joules.

2. \(W = F \times D\)

\D = \frac{W}{F}\)

Plug in what is given: D = 15 Joules/7 Newtons = 2.14 meters.

3. W = − pΔV

ΔV = Vfinal - VInitial = 5 L - 2 L = 3 L

Convert 750 mmHg to atm: 750 mmHg * 1/760 (atm/mmHg) = 0.9868 atm.

W = − pΔV = -(.9868 atm)(3 Liters) = -2.96 L atm.

4. First we must find the final volume using the idela gas law: pv = nRT or v = (nRT)/P = [(.54 moles)(.082057(L atm)/ (mol K))(303K)] / (1.3 atm) = 10.33 L

ΔV = Vfinal - Vinitial = 10.3 Liters - 8 Liters = 2.3 Liters

W = − pΔV = - (1.3 atm)(2.3 Liters) = -3 L atm.

5. \(W = - p * ΔV\) = - 1.8 atm * ΔV.

Given \(p_1\),\(V_1\), and \(p_2\), find \(V_2\): \(p_1V_1=p_2V_2\) (at constant \(T\) and \(n\))

\(V_2= (V_1* P_1) / P_2\) = (1.56 L * 1.7 atm) / 1.8 atm = 1.47 L

Now, \(ΔV = V_2 - V_1=1.47 L - 1.56 L = -0.09\)

W = - (1.8 atm) * (-0.09 L) = 0.162 L atm.

- Gasparro, Frances P. "Remembering the sign conventions for q and w in deltaU = q - w." J. Chem. Educ. 1976: 53, 389.
- Koubek, E. "PV work demonstration (TD)." J. Chem. Educ. 1980: 57, 374. '
- http://en.wikipedia.org/wiki/Work_%2...rmodynamics%29
- http://www.school-for-champions.com/science/work.htm
- http://www.mhi-inc.com/Converter/Energy%20Converter.htm

- Petrucci, et al. General Chemistry: Principles & Modern Applications: AIE (Hardcover). Upper Saddle River: Pearson/Prentice Hall, 2007.
- Serway, Raymond A.; Jewett, John W. (2004).
*Physics for Scientists and Engineers*(6th ed. ed.). Brooks/Cole. ISBN 0-534-40842-7.

- Avneet Kahlon (UCD)
- Gary Rossetto (UCD)

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