Ideal Gas Cycle: N2 Undergoing abcd w/ Varying Pressures

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SUMMARY

The discussion focuses on the ideal gas cycle involving two moles of nitrogen (N2) gas undergoing a cycle labeled abcd with varying pressures. The pressures at states A and D are 5500 Pa, while states B and C have pressures of 1500 Pa. The volumes at states A and B are 1.80 m³, and at states C and D are 8.70 m³. The total work done on or by the gas during the complete cycle was incorrectly calculated by the user, who initially treated the processes A->B and C->D as isochoric, leading to an incomplete understanding of the work done.

PREREQUISITES
  • Understanding of ideal gas laws and behavior
  • Knowledge of thermodynamic processes, specifically isochoric and isobaric processes
  • Familiarity with work calculations in thermodynamics, particularly W = p(deltaV)
  • Basic grasp of pressure and volume relationships in gas cycles
NEXT STEPS
  • Study the derivation of work done in thermodynamic cycles, focusing on both isochoric and isobaric processes
  • Learn about the First Law of Thermodynamics and its application in gas cycles
  • Explore detailed examples of ideal gas cycles, including calculations for work done
  • Review the concept of state functions and path functions in thermodynamics
USEFUL FOR

This discussion is beneficial for students and professionals in physics and engineering, particularly those studying thermodynamics, gas laws, and energy transfer in ideal gas systems.

Luis2101
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Two moles of N2 gas undergo the cycle abcd. The pressure of the gas in each state is
pa = pd = 5500 Pa;

pb = pc = 1500 Pa.


Note that 5500 Pa > 1500 Pa. The volume of the gas in each state is
Va = Vb = 1.80m^3 ;

Vc = Vd = 8.70m^3.

Note that 8.70 m^3 > 1.80 m^3. The gas may be treated as ideal.

--------------------
Part A asks: Find the magnitude of the total work done on (or by) the gas in the complete cycle.

I tried to approach the problem in the following manner and got an incorrect answer for Part A.

I treated the process from A->B, as well as from C->D as Isochoric since Volume is constant, and said no work was done.

I then calculated the work from B->C:
Using: W = p(deltaV) = 1500 (8.7-1.8) = 10,350J.

I said this was the total work done, but it's not...
Any help would be greatly appreciated.

-Luis
 
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