thermodynamics: work

[puffylover]

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is work only done with the volume of the container containing the gas changes? can work occur without change in volume?

 

[UC pre med]

Well it depends what you are talking about because you can do mechanical work and increase the temperature of something. Such as stirring water with a wooden spoon that is the same temperature as the water, so by doing work you are adding thermal energy to it, but since there is no change in volume, you are not doing any pressure-volume work.

Pressure-volume work is the the type of work that is discussed in the first law of thermo U = Q - W, so in this sense, if you do not have a change in volume, then no pressure-volume work is being done, while other types of work may be happening.

Hope this helped, pm if you need further clarification.

 

[dingyibvs]

is work only done with the volume of the container containing the gas changes? can work occur without change in volume?

Short answer: No

Long answer:

No, but don't confuse work with energy. Work is only one type of energy, so while you can't do work without changing the volume, you can in fact change the energy of the system(by heating it, for example).

 

[addo]

Helmholtz free energy measures the amount of useful work obtainable from a closed system at constant temperature and volume. For example, reactions in the cell occur at constant volume, and those reactions are used to do useful work.
Gibbs free energy, the one we are familiar with measures amount of useful work extractable from a closed system at constant temp and pressure.
The long answer is yes, but as long as you are talking about pressure-volume work, the answer is no.
And obviously you dont need to know helmholtz free energy.

 

[puravida85]

In a gas chamber with say a movable piston:


F = P * A

Because, W = F * d

W = P * A * d

We know, A * d = V

Therefore, W = P * V

Case in point, we need volume to change for work to be done ( "delta d" is really meant when saying "d" in W = F * d). Don't worry about anything else really because if work isn't being done, than it must be heat change (q). In the end, it's either going to be change in heat or change in work because E = q - W.

 

[askamsky51]

In a gas chamber with say a movable piston:


F = P * A

Because, W = F * d

W = P * A * d

We know, A * d = V

Therefore, W = P * V

Case in point, we need volume to change for work to be done ( "delta d" is really meant when saying "d" in W = F * d). Don't worry about anything else really because if work isn't being done, than it must be heat change (q). In the end, it's either going to be change in heat or change in work because E = q - W.

Best answer, IMO.

It would be wise to make a graph to better represent a visualization of this case; using pressure on the y-axis and volume on the x-axis. By graphing two points you can find that the work done by the gas is the area underneath the two points... This being said, the graph which has no change in volume is a vertical line on the y-axis (thus no "area" to demonstrate work).

side note: on this graph temperature will increase up and to the right, however keeping in mind that it won't be linear (unless it's an isotherm).