Thermo cycles with only T and mol

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SUMMARY

The discussion focuses on calculating the efficiency of a three-step thermodynamic cycle involving 3.8 mol of an ideal diatomic gas. The cycle consists of a constant volume heating from 210 K to 550 K, an isothermal expansion, and a constant pressure contraction back to the original volume. Key equations used include Q=nCvΔT for the constant volume step and W=nRTln(V2/V1) for the isothermal expansion. The challenge lies in determining the work and heat flow without specific initial or final pressure or volume values, suggesting the use of variables that may cancel out in the calculations.

PREREQUISITES
  • Understanding of the ideal gas law and its applications
  • Familiarity with thermodynamic processes, particularly isothermal and constant volume processes
  • Knowledge of specific heat capacities for diatomic gases (Cv and Cp)
  • Ability to manipulate equations involving work and heat in thermodynamic cycles
NEXT STEPS
  • Study the derivation and application of the ideal gas law in thermodynamic cycles
  • Learn about the efficiency calculations for various thermodynamic cycles
  • Explore the implications of using variables in thermodynamic equations and how they can simplify calculations
  • Investigate the properties of diatomic gases and their specific heat capacities in detail
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, particularly those focusing on ideal gas behavior and efficiency calculations in thermodynamic cycles.

bmartz
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Homework Statement


A three-step cycle is undergone by 3.8 mol of an ideal diatomic gas:
(1) the temperature of the gas is increased from 210 K to 550 K at constant volume;
(2) the gas is then isothermally expanded to its original pressure;
(3) the gas is then contracted at constant pressure back to its original volume.
Throughout the cycle, the molecules rotate but do not oscillate. What is the efficiency of the cycle?


Homework Equations


Q=nCvΔT
Q=nCpΔT
Cv=5/2 R
Cp=7/2 R
W = nRTln(V2/V1)
ε=Wnet/QH

The Attempt at a Solution


I know I need to find the total work and heat flow in. For step one it is easy enough, W=0 and Q=nCvΔT.
I'm stuck at step 2. I know that since it is isothermal ΔEint is 0, so Q=W. I am trying to calculate by:
W = nRTln(V2/V1)
But without any initial or final pressure or volume values I am lost.
 
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Assume the problem as stated gives all the required information. Can you just let either the initial pressure or volume equal to an unknown with the hope that it will cancel out in the end? Just a thought.
 

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