Work by an ideal gas in a thermodynamic cycle

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SUMMARY

The discussion focuses on calculating the work done by an ideal gas during a thermodynamic cycle involving isothermal expansion, isobaric compression, and isochoric processes. The cycle consists of an isothermal expansion from point A to B at 700 K and 9 atm, followed by isobaric compression from B to C at 4 atm. The work done during the isothermal expansion is calculated as 93.16 J using the formula W=nRT*ln(v2/v1). The participants seek assistance in determining the work done during the isobaric compression from B to C, emphasizing the need for pressure and volume calculations at point C.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV=nRT)
  • Knowledge of thermodynamic processes (isothermal, isobaric, isochoric)
  • Familiarity with work calculations in thermodynamics (W=pdV)
  • Ability to manipulate logarithmic functions for volume ratios
NEXT STEPS
  • Calculate the pressure and volume at point C using the Ideal Gas Law.
  • Learn how to derive work done during isobaric processes.
  • Explore the relationship between pressure, volume, and work in thermodynamic cycles.
  • Review the concept of area under the curve in PV diagrams for work calculations.
USEFUL FOR

Students studying thermodynamics, physics enthusiasts, and anyone involved in engineering applications related to gas laws and thermodynamic cycles.

merbear
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1. Homework Statement

Two moles of an ideal gas are carried around the thermodynamic cycle shown in Fig. 18-29. The cycle consists of (1) an isothermal expansion A to B at a temperature of 700 K, with the pressure at A given by pA = 9 atm; (2) an isobaric compression B to C at PC = 4 atm; and (3) an isochoric pressure increase C A. What work is done by the gas per cycle? (See attached picture)

2. Homework Equations

pv=nRT

W=pdv

P=nRT/V


3. The Attempt at a Solution

W=nRT*ln(v2/v1)

For Pa:
9 atm= [2 mol*(.08205784 L*atm/k*mol)*700K]/V2

For Pb:

4 atm= [2 mol*(.08205784 L*atm/k*mol)*700K]/V1

V2/V1= 4/9

W=nRT(ln(4/9)) = 93.16 J (for A to B)

Here is where I got stuck. I can't seem to find the work for B to C without having variables in the answer. Once I can find the work from B to C I can add the work quantities to find the total (and ignore the value for C to B because the total work with no volume change is zero).

Any help in finding work from B to C would be appreciated.

Thank you
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Attachments

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What is the pressure at C? What is the volume at C?
 
You can use PV=nRT to find the volume at point A and C
Then, pluge into the formula W=nRT*ln(V2/V1)
Final subtract it with the area below that triangla W - 3*101325*(V2-V1)
 

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