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29 March 2010

First Law of Thermodynamics and the Sign of Each Term

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

ΔU = Q+W

where each variable has the following meaning:

ΔU 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:

ΔU: Read the axes of the graph, and ΔU = (3/2)PV.
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 ΔU = (3/2)PV, ΔU 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 ΔU and W and plug in.

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

Start with ΔU.  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.  ΔU = Q+W.  Solving for Q, we find that Q = ΔU - W.  Since ΔU 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...

1 comment:

  1. Great job presenting this information. Very useful!