PV Graph Question: Internal Energy Not Zero?

Thread starter bluesteels
Start date Feb 2, 2022
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The discussion centers on the confusion regarding the internal energy of a gas in a PV graph, particularly in the context of a process from state a to state b. It is noted that while the gas returns to its original state, the internal energy is not necessarily zero for the specific process being analyzed. The focus is on understanding that internal energy depends on the state of the system, not just the path taken. Clarification is provided that the question pertains only to the process from a to b, which may involve changes in energy despite the gas returning to its original state. This emphasizes the importance of analyzing the specific process rather than assuming zero internal energy based solely on the initial and final states.

Feb 2, 2022

bluesteels

Homework Statement: The idealized cycle shown is known as the Otto cycle. (Figure 1)Suppose an engine is executing this Otto cycle, using a gas (not necessarily ideal) as its working substance. From state A to state B, the gas is allowed to expand adiabatically. (An adiabatic process is one in which no heat is added to, or given off by, the working gas.) The gas is then cooled at constant volume until it reaches state C, at which point it is adiabatically compressed to state D. Finally, it is heated at constant volume until it returns to state A.

Relevant Equations: adiabactic delta u = -w

isothermal Q=W

Figure 1:

Screen Shot 2022-02-03 at 12.04.03 AM.png

im so confused on why is the internal energy not zero for the 2nd picture because i thought if the gas returning to its original states so it zero

Screen Shot 2022-02-03 at 12.02.43 AM.png

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Feb 2, 2022

haruspex

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bluesteels said:

i thought if the gas returning to its original states so it zero

The question only asks about process ##a\rightarrow b##.

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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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