The first law of thermodynamics is a statement of energy conservation. In equation form, it states that

Δ

*U = Q+W*where each variable has the following meaning:

Δ

*U*is the change in the gas's internal energy

*Q*is the heat ADDED to the gas

*W*is the work done ON the gas.

A typical test question will show a pressure-vs-volume graph, and expect the student to use the graph to determine a value, or at least a sign, for each of the quantities. Therefore, it's worth memorizing the method of determining each variable from a PV diagram:

Δ

*U*: Read the axes of the graph, and Δ

*U*= (3/2)

*PV.*

*Q*:

**CANNOT BE FOUND FROM THE GRAPH!**

*W*: Look at the area under the graph.

The quiz/poll thingie I had up last week is shown to the right. This question only asks for the signs of the three variables.

Based on the definition Δ

*U*= (3/2)

*PV*, Δ

*U*is positive when the gas's temperature increases, and negative when the temperature drops. (Why? Look at the equation, and remember that

*PV = nRT*.)

The sign of

*W*can be trickier. Work is done ON the gas when the gas's volume DECREASES. Think of a piston compressing the gas... a force is applied ON the gas over some distance. So we consider

*W*to be a positive quantity when a gas compresses, and a negative quantity when a gas expands.

And finaly, the sign of

*Q*can only be determined from the first law equation. One must find Δ

*U*and

*W*and plug in.

And now, the answer to this week's quiz:

Start with Δ

*U*. This process starts and ends at the same spot on the PV diagram -- the product of PV does not change. Therefore, ΔU is

**zero**--

**there is no net change in internal energy.**

Now look at

*W*. Consider each sub-process individually. In process A-B, the gas expands, so

*W*is negative. In process B-C, there is no volume change, so no work is done on or by the gas. And in process C-A, the gas's volume decreases, so

*W*is positive. But what is the sign of

*W*for the overall process A-B-C-A? There is more area under the graph in the expansion from A-B than in the compression from C-A. So, the net value of

*W*is

**negative -- net work is done BY the gas.**

And finally, use the first law to find the sign of

*Q*. Δ

*U = Q+W.*Solving for

*Q*, we find that

*Q =*Δ

*U - W*. Since Δ

*U*was zero and

*W*was negative, then algebraically,

*Q*must be positive. This means that, in net,

**heat is added to the gas.**

As a side note -- this process represents a cycle of a heat engine. Despite what happens in each individual process, IN NET, some amount of added heat is converted into some amount of work done by a gas. That's what is meant by a heat engine.

More polls soon...

GCJ

Great job presenting this information. Very useful!

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