First law of thermodynamics and gaining heat

[garytse86]

This is about the system and the surroundings.

I am not sure whether the signs are correct:

When the system is gaining heat, the sign is positive.
When the system is losing heat, the sign is negative.

When work is done on the system, the sign is negative.
When work is done by the system, the sign is positive.



When a syringe expands, is the work positive?


I am really having problems with the positive and negative signs because they differ with textbooks. What I thought was that the signs of heat and work are always opposite, as I stated above.

Gary

 

[Bystander]

The internal energy, U or E, is defined as q - w in most engineering and physics texts, and as q + w in many chemistry texts. Sticking to PdV work for the moment, the first case defines w as "work done by the system;" chemists define w as "work done ON the system." In the first case w is the integral of PdV and positive when volume increases; in the second, w is minus the integral and negative when V increases. Both cases sum to the same internal energy. The first case is appropriate historically, and for engineering since interest is in "work" done by a system for a given heat input. The second case is more convenient for chemical reasoning, since it emphasizes work as being done on a system to increase its internal energy.

When a syringe expands, w is positive if the system is defined as the gas contained in the syringe, and the expansion is done against some external force, and you are working with the first definition of internal energy.

 

[Chi Meson]

Just two years ago, the AP people decided to change the definition of positive work to follow the traditioanl "chemistry" viewpoint. THis caused five or six textbooks (at least) to change their definitions. It seems as though the definition is leaning toward "work done on system" is positive.

Either way, the first law is consistent: since work and heat are two ways of transferring energy (excluding radiation), the change in energy contained by a system of matter will be the sum of the heat that goes in plus the work done on it.

Regarding the expanding syringe, since the plunger is being pushed, a force is applied to it and it moves through a distance; therefore, work is done on the plunger by the gas. THe gas cannot do work and still retain the original amount of energy. So either way you define work, if the gas does the work on something outside the system, the system loses internal energy.

 

[turin]

Originally posted by garytse86
This is about the system and the surroundings.

I am not sure whether the signs are correct ...

It doesn't matter, as long as you are consistent. We cannot answer your question directly unless you write the first law that you are using (there is only one first law, but I mean, "write the equation so we can see what you're adding and what you're subtracting.") The law refers to a system, so ΔU is the change in internal energy of the system regardless of convention (you just have to know what you're defining to be the system). Then, to really overstep my bounds and generalize the first law:

ΔU = "the influx of energy"
(This is not a definition of ΔU; this is a statement of a non-trivial physical law.)


There are then four ways to write it as there are four ways to choose the conventions:

1) W is work done to the system, and
Q is heat tranfered into the system:

ΔU = W + Q

2) W is work done to the system, and
Q is heat tranfered out from the system:

ΔU = W - Q

3) W is work done by the system, and
Q is heat tranfered into the system:

ΔU = Q - W

4) W is work done by the system, and
Q is heat tranfered out from the system:

ΔU = -W - Q

First, you must choose your preferred convention and corresponding first law equation. Then, the values that you plug in for W and Q are positive if they correspond to the conventional definition, and negative if they correspond to the complementary definition.


For some examples, assume that we have chosen convention 3). This tells us that the first law is to be written:

ΔU = Q - W

Let's say I have a canister with rigid thermally conductinvg walls (like a sealed tin can) that contains a gas at a high temperature. Then, I can use the conventional definitions to set:

W = 0, and Q = negative value (the complement of "into" is "out from").

Then, ΔU is negative over time. Physically, this means that the gas in the can will lose energy over time. This result is independent of the convention that we use, an important clue.

Let's say I have a perfectly insulated cylinder containing a gas at high temperature with a perfectly insulating massive piston on top that is being forced to compress the gas. Then, I let go of the piston and it flies upwards. The corresponding values are:

W = possitive value, and Q = 0.

Then, ΔU is again negative over time. Again, physically, this means that the gas in the can will lose energy over time, and this result is again independent of the convention that we use.