## Heat Engines

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Part 1:

1.   1/P= 6.31 x 10^4   ,   VFlask= 157.2 mL

2.   V=6.31x10^4(1/P)- 157.2

3.   B= 6.31x10^4(1/P) – 157.2 – v

Four Process Cycle:

 Step Process Q W Einternal 1 Isochoric + 0 + 2 Isothermal + - 0 3 Isochoric - 0 - 4 Isothermal - + 0

2. The area under the PV curve is the work done in there system.  The sign of the work determines whether work was done on or by the system

3. The internal energy in the system remains constant during the isothermal stage because the temperature of the whole system does change during the process

4.  WNet= 374.2mL*kPa x (1m3/1000000cm3) = 3.742x10^-4 J

Three Process Cycle:

 Step Process Q W EInternal 1 Isochoric + 0 + 2 Isothermal + - 0 3 Isobaric - + -

2. We kept the temperature decreasing while we are decreasing the volume of the cylinder allowing the temperature to stay constant.  PV=nR(T)

Extension Questions:

1. The net work done by the process would be the exact same as the first cycle because their area inside would be the same, the only difference would be the sign of the work done.
2. The Carnot cycle differs from the four-step process because in the Carnot cycle, each process is reversible and all of the heat is transformed to work put out by the system.

 Step Process Q W Einternal 1 Isothermal + - 0 2 Adiabatic 0 + + 3 Isothermal - + 0 4 Adiabatic 0 - -

Part 2 Graph:

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