Hello and thank you for your question.
First some background:
Take a look at the illustrations on this page:
Engine Cycles
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/engcyc.html#c1
As you see, the vertical axis on your graph is Pressure, and the
horizontal axis is Volume.
So if you'll imagine a piston cycling in-and-out of a cylinder, you
can see that the volume gets larger when the piston is pushed out of
the cylinder, and smaller when the piston moves back in.
So when the gas is in the process of expanding (that's happening from
point A to point B ("portion AB") because the volume is increasing),
work is being done <by> the gas--it's pushing the piston out. That's
when heat is being added to the system--the heat provides the energy
to do the work.
Next comes portion BC. Here the volume is constant but the pressure
is falling. Heat now is being extracted but no work is being done
(you see, volume being constant means the piston isn't moving). The
pressure falls when the molecules decrease their speed and kinetic
energy--they're getting cooler as heat leaves the system)
Next, in portion CD, the gas is being compressed (the volume is
decreasing), so work is being done <on> the gas, and heat continues
to be extracted from the system.
Finally (well, not really finally because this 4-step process
continues indefinitely), in portion DA, the volume is constant (so no
work is being done) but the pressure is increasing with the addition
of heat to the system.
Now, take a look at
Thermo Example 4
http://users.ipfw.edu/maloney/thermoex4.html
It provides a slightly more complicated graph (a trapezoid instead of
a rectangle) but as that page explains, your problem is easier because
it's isobaric (constant pressure when the work is being done).
"The work done by, or on, an ideal gas is computed by taking the
product of the pressure and the change in volume. That is the general
prescription for determining the work, but this prescription can only
be applied directly to isobaric processes. For other types of
processes, the calculation is more involved."
Adapting what follows on that page to the numbers in your graph, the
pressure is constant (that is what isobaric means) for process AB.
So in portion AB the volume changes from [1.5 to 4] x10negative6 m3 so
the work done by the gas is the product of that difference [2.5 x10^-6
(m3)] times the constant pressure [4 X 10^5 (N/msquare2) = one joule
[that's what you typed out as "IJ" in part (a) of your question]
here's the arithmetic:
2.5 * 10^-6 * 4.0 * 10^5
= 10 * 10^-1
= 1 joule
(the units check because m3 * N/m^2 = mN which is the definition of joules)
As described above, work is zero from BC and from DA
In portion CD there's a negative sign because work is being done <on>
the gas and the amount of work is the same 2.5 * 10^-6 but times the
lower pressure
2.0 * 10^5 = -0.5 joule
That's it!
Search terms used:
"P-V diagram"
Rectangle work "PV diagram"
If you find any of this unclear, please request clarification. I
would appreciate it if you would hold off on rating my answer until I
have a chance to respond.
Sincerely,
Richard-ga |
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