Work done by Octane Combustion Engine

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SUMMARY

The discussion centers on the calculations related to an insulated internal combustion engine fueled by one mole of octane at a constant pressure of 5.0 bar and an initial temperature of 200°C. The final temperature was determined to be 2662.42 K, with the heat exchanged and the change in enthalpy (Δh) both calculated as zero. The work performed by the gas was calculated using the formula w = -PΔV, resulting in -1233280.344 J. The change in internal energy (Δu) was derived from the relationship Δu = Δh - PΔV.

PREREQUISITES
  • Understanding of the ideal gas law (PV = nRT)
  • Knowledge of thermodynamic equations for work (w = -PΔV)
  • Familiarity with enthalpy changes (Δh) and internal energy (Δu)
  • Basic principles of combustion reactions and stoichiometry
NEXT STEPS
  • Explore the derivation and application of the ideal gas law in thermodynamic processes
  • Study the relationship between enthalpy and internal energy in closed systems
  • Learn about the thermodynamic properties of octane and its combustion products
  • Investigate the implications of constant pressure processes in thermodynamics
USEFUL FOR

Students in thermodynamics, chemical engineers, and anyone involved in combustion engine analysis or energy conversion processes.

SniffLimit
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Homework Statement



There is an insulated internal combustion engine fueled by one mole of octane. It operates at a constant pressure of 5.0 bar. The initial temperature is 200°C. Calculate the heat exchanged with the surroundings, the Δh for the gas, the final temperature, the work performed by the gas, and and the Δu for the gas.

I'm given the balanced equation:

C8H18 + 12.5 O2 + 50 N2 → 9 H2O + 8 CO2 + 50 N2

I'm also given the enthalpies of formation and the CP (at 298 K) for each product and reactant, but I've already successfully found the final temperature, so I don't think it's necessary to include them here.

I just need to find the work and the Δu at this point.

Homework Equations



w = nRΔT
w = -PΔV
Δu = q + w
PV = nRT

The Attempt at a Solution



The heat exchanged with the surroundings is 0. The Δh for the gas is 0. The final temperature is 2662.42 K. I've confirmed that these three answers are correct. To find the work, I think I would just use the work equation above, but I'm not sure what it wants me to plug in for n. Moles of reactants? Moles of products?

I suppose I could use the ideal gas equation to find the volumes at the initial and final temperatures, and then use the other work equation.

(500,000 Pa) V = 63.5 (8.314) (473.15)
Vi = 0.4996 m3

(500,000 Pa) V = 67 (8.314) (2662.43)
Vf = 2.9661 m3

ΔV = 2.466560689 m3

w = (-500,000 Pa) (2.466560689 m3)
w = -1233280.344 J

But how would I go about finding the Δu?

Δu = Δh - PΔV

Whoops, I solved it...
 
Last edited:
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SniffLimit said:
Whoops, I solved it...
LOL. Welcome to the PF! :smile:
 
Yeah, sometimes just organizing your thoughts to post the question helps.

Feel free to try again next time you have problems with a question :wink:
 

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