Thermodynamics Cycle: Gas Temperature and Work Calculation | Homework Solution

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SUMMARY

The discussion focuses on solving a thermodynamics cycle involving a gas subjected to heating, cooling, and isothermal compression. The calculations yield temperatures at points A, B, and C as 64.5K and 129K, with work done on the gas calculated as 653J and heat transferred as 54J for the entire cycle. Key equations utilized include the first law of thermodynamics (ΔU = Q + W) and the ideal gas law (PV = nRT). The analysis highlights the importance of knowing the number of moles and the type of gas for accurate calculations.

PREREQUISITES
  • Understanding of the first law of thermodynamics (ΔU = Q + W)
  • Familiarity with the ideal gas law (PV = nRT)
  • Knowledge of specific heat capacities (Cp and Cv)
  • Basic principles of thermodynamic cycles
NEXT STEPS
  • Learn about calculating work done in isothermal processes
  • Study the implications of state functions in thermodynamics
  • Explore the differences between monatomic and diatomic gases
  • Investigate the significance of heat transfer in thermodynamic cycles
USEFUL FOR

Students studying thermodynamics, engineers working with gas systems, and anyone involved in thermal energy calculations will benefit from this discussion.

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Homework Statement


I.The gas is heated at constant pressure until the volume is 4.4L at point B
II. The gas is cooled at constant volume until the pressure decreases to 1.2 atm
III. The gas undergoes an isothermal compression back to point A

a) Find the temperature at Point A,B,C
b) Find the work done on the gas ,the heat transferred to the gas and the change for each process
c) Find W,Q and delta U for the whole cycle

Homework Equations


[/B]
delta U=Q+W
PV=nRT
1 atm=1.013x10^5 N/m^2
Cp=5/3R+R

The Attempt at a Solution


a)
pv=nRT
Tb=PbVb/nR
Tb=2.4*4.4/(1*8.31) * 0.001* 1.013*10^5
=129K
Ta=129K/2
=64.5K
Tc=129K/2
=64.5K

b)
Wab=pv
=2.4*(4.4-2.2)* 1.013x10^5 *0.001
=535J
Qab=n*Cp*deltaT
=0.32* 8/3*R *(129-64.5K)
=457J
U=Q+W
delta Uab=457+535
=992J

Wbc=0
Since the volume keep unchange
Qbc=n*Cv*deltaT
=0.32*5/3*R*(64.5K-129K)
=-285J
U=Q+W
U=-285J

U=0
Since it is isothermic process
Wca=nRTc *ln(Vc/Va)
=0.32*8.31*64.5K*ln(4.4/2.2)
=118
Q+W=0
Q=-118J

c)
W=Wab+Wbc+Wca
=535+0+118
=653J
Q=Qab+Qbc+Qca
=457-285-118
=54J
deltaU=992-285+0
=707J
 

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Some comments:
1. How many moles of gas are there? You seem to be using 1 and .32 in different places. There has to be more information provided in order to determine this.
2. What is the ##\Delta U## for one complete cycle? (hint: is U a state function?).
3. What kind of gas is this? (e.g monatomic, diatomic etc. Are you given this or given the Cv or Cp?).

AM
 

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