Let’s talk about moving boundary work. Imagine we have a piston-cylinder device like this,
or ones similar to automobile engines. The gas inside the cylinder can expand or compress.
If we draw a boundary like this to include the gas, and we allow the gas to expand, our boundary
expands along with it. This is a moving boundary. Now when this expansion or compression occurs,
there is work involved. This work is called moving boundary work. We can analyze the moving boundary
work assuming that the system remains pretty much in equilibrium. Hopefully, you remember that work
is force times distance. If our piston movies up a tiny amount like this, then the amount it moved up
is the differential amount ds. So we can say the a differential amount of work, in simple terms,
a tiny amount of work, is equal to the force times that tiny amount ds. Well, force, is pressure of
the gas times the cross-sectional area of the piston so we can write it like this. On a side
note, the pressure of the gas is actually the pressure at the inner surface of the piston.
Now going one step further, instead of saying the cross-sectional area times ds, we can actually
just use differential volume. So volume change, dV, is positive during an expansion process,
so when the volume increases, and negative in a compression process, when volume decreases.
Now we obviously can’t just think about tiny amounts of work,
we need the full picture, so we definitely need to integrate, in other words, add up all those
tiny amounts of differential works together. So we can write our equation like this. The boundaries
of the integral are the initial and final volume values. The units would be kilojoules.
(kJ) To explain this better,
let’s draw a pressure vs volume graph. We will also show the piston like this.
Now here is our initial point. So we have high pressure, but low volume. Now our piston moves
likes this, and stops at the second stage. Now our volume increased, and the pressure dropped. This
line here, is called the process path. It tells us the pressure at any volume, and vice versa.
The work done is then the area under this line. Boundary work is zero whenever the volume stays
constant. As shown on a graph like this, there is no boundary work done during the process.
One last thing we need to talk about is called a polytropic process.
When gases expand and compress, pressure and volume can be related using this equation.
If a process obeys this equation, then it’s a polytropic process. Here, C and n are constants.
N is also called the polytropic index. There are a few other equations that are related to this
process. For example, to find the boundary work when it comes to a polytropic process,
we can use this equation. The constant C can be found using this equation. This part here can
be used to compare initial pressure and volume to final pressure and volume and to find the n
constant. Now let’s get to some examples to see how we can use the equations we talked about.
In this question, we have helium in a piston-cylinder and it’s compressed.
We need to find the initial temperature, final temperature
and the work required to compress the helium. Let’s write down what we know.
The initial volume is 5 cubic meters. There is 1 kg of helium.
The pressure at the initial and final state is 180 kPa since it’s maintained to be constant.
And the final volume is 2 cubic meters. We also need the gas constant of helium,
which we can find using our table. That’s the value right here.
Now let’s find the initial temperature. To do so, let’s write our ideal gas equation. Since
we’re looking at everything at the initial state, I will use a subscript one for the variables.
Let’s plug in our values. Now we can solve for the initial temperature.
Next, we can find the final temperature. We will use the same equation as before.
This time, we will use the values at the final state. Remember that the pressure stays the same.
Let’s solve for the final temperature. The last step is to figure out how much
work was required to compress the helium. Let’s draw a pressure vs volume graph. So the
initial state starts right here, with a volume of 5 cubic meters and a pressure of 180 kPa.
The pressure stays constant, which means it’s a straight line to right here, where our final
volume is 2 cubic meters. To find the work, all we need is find the area under the graph. So
that’s 180 multiplied by 2 minus 5. We also need a unity conversion ratio since we want our answer in
kilojoules. We got a negative answer, which simply means that this is a compression process, in other
words, volume is decreasing and it’s a work input. In this question, we have a piston-cylinder device
with nitrogen gas. We need to find the boundary work done when the nitrogen expands through a
polytopic process. First, let’s write down what we know. The initial pressure is 130 kPa, the
temperature is 120 degrees Celsius which is 393.15 Kelvin, and the volume is 0.07 cubic meters.
After the expansion, the pressure is 100 kPa, and the temperature is 100 degrees Celsius which is
373.15 Kelvin. Now let’s write down the equation for work done during a polytropic process. Looking
at the equation, we need to figure out 2 unknown values: the 2nd state volume, and the n constant.
To figure out the second state volume, we need to use the ideal gas equation that compares 2 states.
Let’s plug in the values we have. Now we can solve for the final volume.
Next, we need to find the n-constant. To do so, we are going to use this equation. Let’s
plug in our values. Now we can go through the process of solving for n. Make sure you remember
exponent rules. Solving gives us the n constant. Now that we have both unknowns, let’s go back
to our equation and plug our values in. Solving gives us the boundary work and that’s our answer.
Let’s take a look at this question where we compress a gas.
We need to figure out the work done during this compression process
by plotting a pressure volume diagram, and by integrating. Let’s write down what we know.
The initial volume is 0.42 cubic meters and the final volume is 0.12 cubic meters.
We are also given the equation for the pressure change, so let’s write that down as well.
We’ll draw the graph first. The 2 points of interest for the volume would be 0.42 and 0.12. So
we just plug in 0.42 into our equation to figure out the pressure. That’s our initial pressure.
Next, we need the final pressure, so let’s plug in the final volume into the equation.
Now we can plot these 2 values on the graph. Here is point 1,
and now point 2. Let’s connect the dots. To find the work, we need to calculate the
area under the graph. We have 2 pieces, a right angle triangle and a rectangle.
The area of a right-angle triangle is ½ times base times height. So the height is P2 minus
P1. The base length is V2-V1. Solving gives us the area of the triangle piece.
Now the bottom part is a rectangle, so it’s just base times height. The height is P1, and the base
is V2 – V1. The total work is then the addition of these 2 pieces. We get a negative value,
which means this is a compression process, and work is being done on the system, so work input.
Next, we need to find the work again, but this time, by integration. Let’s write our equation.
Now we plug in the equation for pressure. The lower bound is initial volume, and upper bound
is the final volume. Let’s solve the integral and as expected, we get the same answer. Again,
it’s negative because it’s a compression process, so this is work input. In simple terms,
we have to input work to compress the gas. That should cover the types of problems you will
face when it comes to moving boundary work. Thanks for watching and best of luck with your studies.