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.
