- #1

TheStebes

- 11

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## Homework Statement

A sample of an ideal gas goes through the process shown below. From A to B, the process is adiabatic; from B to C, it is isobaric with 98 kJ of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 158 kJ of energy leaving the system by heat. Determine the difference in internal energy, Eint, B-Eint, A.

(sorry I can't post the link to the actual graph)...

Point B: P=3atm, V=.09 m^3

Point C: P=3atm, V=.40

Point D: P=1atm, V=1.2

Point A: P=1atm, V=.20

## Homework Equations

[tex]\Delta[/tex]E= Q + W

For cyclic process, [tex]\Delta[/tex]E=0, Q = -W

isobaric process, W=-P(V[itex]_{f}[/itex]-V[itex]_{i}[/itex]

isothermal process, W=nrTln(V[itex]_{i}[/itex]/V[itex]_{f}[/itex])

## The Attempt at a Solution

I understand that based on conservation of energy, after completing one complete cycle, the overall change in energy will be 0. Basically, this means the net work done equals the area enclosed by the path. I'm confused at how you would solve for the energy at a single point though -- in order to find the difference between E,B and E,A. What equation would I use to calculate such an energy?

Using the equation Q=-W:

98 - 158 = -[W[itex]_{B,C}[/itex] + W[itex]_{C,D}[/itex] + W[itex]_{D,A}[/itex] + W[itex]_{A,B}[/itex]]

W[itex]_{B,C}[/itex] = -P(V[itex]_{f}[/itex]-V[itex]_{i}[/itex] = -94kJ

W[itex]_{D,A}[/itex] = -P(V[itex]_{f}[/itex]-V[itex]_{i}[/itex] = 101.325

W[itex]_{C,D}[/itex] and W[itex]_{A,B}[/itex] are isothermal processes which can be solved with W=nrTln(V[itex]_{i}[/itex]/V[itex]_{f}[/itex]), but no information is provided about the temperature.

At this point, I'm stuck.

Any help would be greatly appreciated.

Scott