Integral Application Question

[Johasua]

So this is an application question for integrals, here's the question:

Suppose that gas in a circular cylinder is being compressed by a piston.
a) If p is the pressure of the gas in pounds per square inch and V is the volume in cubic inches, show that the work done in compressing the gas from state (p1,V1) to state (p2,V2) is given by the equation in the picture.




I really have no idea on how to solve this question. I haven't taken physics before so these questions are really confusing me

 

[tkhunny]

I am going to beat you up a little bit. You said the magic words!

I do not believe you have no idea. Never say that again. Never think it again.

You do not need a course in physics. You need some thinking time.

Look at that piston.
If you had to compress it it a little, would it be very hard?
If you had to compress it more, would it be harder or easier?
Have you EVER met an internal combustion engine? That piston compresses gas. How does it work. In particular, how does a deisel work? Very high compression and you get an explosion!! It's awesome.

After a little thinking, instead of throwing hands in the air and giving up, we conclude that it gets harder and harder to move the piston. The more we move it, the harder it gets.

Now comes the interesting part. You've been studying integrals!! There MUST be a connection or your book or teacher are a little crazy!

We have a standard gas law, don't we? Did the problem give you this information? If not, it had better be pretty common knowledge. If so, use it!

According to the "Ideal Gas Law", we have [(Absolute Pressure) * Volume / (Absolute Temperature) is Constant]. Do we need this or did the problem give us something else?

Please have more than "no idea" and give us a clue where you are in your study of integrals.

 

[Johasua]

In part b it does say the p and V obey the gas law pV^1.4 = constant. I know that work = the integral (over a certain distance) of the Force, but how would I apply that here?

thanks

 

[galactus]

We assume that the pressure and temperature of the gas are the same at every point inside the cylinder, so the gas is in equilibrium.

The force due to gas pressure on the piston is $\displaystyle F=P\cdot A$

where A is the area of the piston cross-section. If the force due to gas pressure is different from the outside force, the piston will move.

If the piston is moved a small distance dy, the change in volume of the gas is $\displaystyle dV=Ady$

and the work done is $\displaystyle \displaystyle dW=F\cdot dy=(P\cdot A)dy=P\cdot (Ady)=PdV$

If the state of the gas changes slowly, so the gas is at equilibrium at each state in between, the work done by the gas is the sum of the components $\displaystyle dW=PdV$.

If the volume is initially at $\displaystyle V_{1}$ and ends at $\displaystyle V_{2}$, the total work can be expressed by the integral:

$\displaystyle \displaystyle W=\int_{V_{1}}^{V_{2}}PdV$

The work done depends not only on the initial and ending states, but also on the states in between.

An ideal gas satisfies $\displaystyle \displaystyle \frac{PV}{T}=\text{constant}$.

T is the absolute temperature in Kelvin.

so the state of an ideal gas can be found from its pressure and volume.

The work done can be expressed as the area under the curve.

adiabatic is when the gas changes state without exchanging heat with its surroundings.

So, in this process, the volume and pressure are related by $\displaystyle PV^{c}=k$

where c and k are constants.

So, we use $\displaystyle \displaystyle PV^{1.4}=k$

Here is an example with expansion.

Say that air in a diesel engine expands from $\displaystyle \displaystyle 1.2 \;\ in^{3}$ to $\displaystyle 12 \;\ in^{3}$.

The initial pressure is $\displaystyle \displaystyle 600 \;\ \frac{lb}{in^{2}}$.

Find the work done.

(** take a look at your problem. It has $\displaystyle \displaystyle 50 \;\ \frac{lb}{in^{3}}$.

I think that should be $\displaystyle \displaystyle 50 \;\ \frac{lb}{in^{2}}$.)



$\displaystyle \displaystyle k=600(1.2)^{1.4}=774.47$

Then, we have $\displaystyle \displaystyle P=774.47V^{-1.4}$

and the work done is $\displaystyle \displaystyle W=\int_{1.2}^{12}774.47V^{-1.4}dV=1083.41 \;\ in-lb$

See what to do now?. Your compressing instead of expanding.

Just for fun, the final pressure for your gas would be $\displaystyle \displaystyle P_{f}=P_{i}\left(\frac{V_{i}}{V_{f}}\right)^{1.4}$

$\displaystyle \displaystyle =50\left(\frac{243}{32}\right)^{1.4}=854.3 \;\ \frac{lb}{in^{2}}$

One would expect the pressure to rise since it is being compressed. Temperature would also increase, but that is not part of the problem.