Isobaric compression from isotherm to adiabat

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SUMMARY

The discussion focuses on determining the pressure and temperature at three points in a thermodynamic process involving 150 mg of helium. The initial conditions include P1 at 3 atm, P2 at 1 atm, and temperatures T1 and T2 both at 971.85 K. The ideal gas law (PV=nRT) is applied to find these values, and the process transitions from isobaric compression to an adiabatic process, utilizing the relationship PVγ = constant, where γ for helium is 5/3. The final pressure P3 can be calculated using the derived relationship from the adiabatic process.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV=nRT)
  • Knowledge of thermodynamic processes, specifically isobaric and adiabatic processes
  • Familiarity with the concept of specific heat ratios (γ) for gases
  • Basic skills in algebra for solving equations
NEXT STEPS
  • Study the derivation and application of the Ideal Gas Law in various thermodynamic scenarios
  • Learn about the implications of isobaric and adiabatic processes in real-world applications
  • Explore the calculation of pressure and temperature changes in adiabatic processes using PVγ = constant
  • Investigate the properties of monatomic gases and their specific heat capacities
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, particularly those focusing on gas laws and processes, as well as professionals in fields such as mechanical engineering and physical chemistry.

nlingraham
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Homework Statement



The figure(Figure 1) shows a thermodynamic process followed by 150 mg of helium. Determine the pressure (in atm) at points 1, 2, and 3. Determine the temperature (in °C) at points 1, 2, and 3.

http://imgur.com/B7lwQ

n=.0375 mol
P1=3 atm
P2=1 atm
T1=971.85 k
T2=971.85 k
V1=1000 cm^3
V2=V3=3000 cm^3

Homework Equations



PV=nRT

The Attempt at a Solution



I'm only on the pressure part right now. From the given chart P1=3 atm. I used PV=nRT, and got that T1 and T2 are 971.85 K. With that, I used ideal gas law again to get that P2=1 atm. However, I don't know how to get the pressure at point 3 since I don't know the temperature at that point. It looks like it goes through an isobaric compression from the isotherm to the adiabat to me. Any help is appreciated.

(303000)(.001)=(.0375)(8.314)T, T1=T2=971.85 K

P(.003)=(.0375)(8.314)(971.85), P2=1 atm
 
Last edited:
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From 1 to 3, the He goes through an adiabatic process such that

PVγ= constant

So you will have to use this to get P3
 
Ok, I get the PVγ, where γ=5/3 since its a monatomic gas, but what exactly is the constant? Is that supposed to be nR?
 
Ok, I think I got it. So it would be P2V2γ=P1V1γ and then just solve for P2
 
Thank you for your help!
 

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