Heat Engine and Cycles

A candle burning the oxygen and suffocating itself.

A candle burning the oxygen that it gets by induction which keeps the flame burning.

Top: Graph of an Isobaric Process where V and T are the only variables, notice how their is a linear correlation.
Bottom Left: Graph of an Isochoric Process where p and T are the only variables, and have a linear relationship.
Bottom Right: Graph of an Isothermal process where p and V are the only variables, and have an inverse relationship.

Now that we understand the relationship of the Isobaric, Isochoric and Isothermal Process, we can set up an equations where we have one pair of variables at time intial equal to those a t-final. By doing this we can just solve for whatever variable we did not measure or have originally!

Notice how we can find the pressure at any given volume, since we were given pressure and volume initial and were told n and T were constant. here we first solved the pressure at 20dm^3 then at 10dm^3.

PV graphs of isobaric, Isocholoric, Adiabaitc, and Isothermal Processes.

We solve for the work of gas by taking the integral of Pdv which gave us P_V_ln(V/V_).

Next we would solve for the W done by gravity which is (3/4)mgh for this specific case since the water tank was only 3/4 full of water. This also means that there is air in the tank, which means their is work being done by the air. The work air exerts is (1/4)*atmospheric pressure * 50. Once we add all of these values we finally get our answer or 1.793 *10^7 MJ.

Here we have a problem where Volume and Pressure are variables. So what we did was take the integral of P(v)dv but substitued nRT/V for P(v) which we can use an integral to describe its change since the only thing changing would be V. From there we can substitute PV back in or leave it as nRT. 

A PV graph was given to us which contained 4 points, the first thing we did was observe when work was being done by the gas(when P or V increase) and when work was being done on the gas(when P or V decrease). Then we went ahead and found the internal energy at each point which we describe to be equal to (3/2)PV.

From there what would be done is to label each of the sides A,B,C,D and let each side represent the change in internal energy which we could solve for by subtracting the final point from the initial point(like vectors). Finally we could find the Work done by using pdelV, then use the first law of thermodynamics to solve for Heat(Q). We could also solve for Worknet by adding the work from each side.

Here we have the efficiency of our system, which can be solved for by taking the ratio of Work over Heat added to our system.