Explain Graphs -- adiabatic Expansion of mono- di- and poly- gas

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SUMMARY

The discussion centers on the mathematical explanation of adiabatic expansion for monoatomic, diatomic, and polyatomic gases, specifically through the differential equation dP/dV = -gammaP/V. Participants clarify the distinction between isothermal and adiabatic processes, noting that adiabatic curves are generally steeper than isothermal curves. The conversation emphasizes the importance of understanding the mathematical solutions to these equations to fully grasp the behavior of gas expansions. Key contributors include users lychette and chestermiller, who provide insights into the mathematical analysis of these processes.

PREREQUISITES
  • Understanding of thermodynamic concepts: isothermal and adiabatic processes
  • Familiarity with the differential equation dP/dV = -gammaP/V
  • Knowledge of homogeneous first order linear ordinary differential equations
  • Basic principles of gas behavior and internal energy changes
NEXT STEPS
  • Study the mathematical solutions to the differential equation dP/dV = -gammaP/V
  • Explore the implications of the First Law of Thermodynamics on adiabatic processes
  • Learn about the specific heat capacities for monoatomic and diatomic gases
  • Investigate graphical representations of isothermal and adiabatic curves
USEFUL FOR

Students and professionals in physics, particularly those studying thermodynamics, as well as anyone interested in the mathematical modeling of gas behaviors during expansion processes.

Shreyans Jain
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"How it can be explained mathematically by dP/dV = -gammaP/V
AdiabaticProcesses_1.PNG
 
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What are your thoughts on this?
 
Chestermiller said:
What are your thoughts on this?
I am not able to figure out which has greater slope kind of confused in negative slope of adiabatic
 
Shreyans Jain said:
I am not able to figure out which has greater slope kind of confused in negative slope of adiabatic
do you understand that ISOTHERMAL means at constant temperature and that ADIABATIC means no heat transfer?
 
Do you know the solution to that differential equation that you wrote down? If so, what is it?
 
The yellow curve shows an isothermal expansion, during the expansion heat energy must be supplied to keep the temperature constant.
The other 3 curves are all adiabatic expansions so no heat energy enters or leaves the gas. Heat energy (internal energy) is removed from the gas and therefore the values of P and V are lower than for the isothermal expansion.
Mathematical analysis will give the details of the curve shapes.
In general adiabatic curves are steeper than isothermal curves...I realize this is not a mathematically rigorous statement but it is helpful ...I hope.
Ps...make sure you are clear about this before you worry about the details of monatomic, diatomic molecules etc !
 
Thanks lychette and chestermiller sir, i know about the processes and i also know what happens when heat is taken out but i was not able to get the results by the differential equation.
 
lychette said:
The yellow curve shows an isothermal expansion, during the expansion heat energy must be supplied to keep the temperature constant.
The other 3 curves are all adiabatic expansions so no heat energy enters or leaves the gas. Heat energy (internal energy) is removed from the gas and therefore the values of P and V are lower than for the isothermal expansion.
Mathematical analysis will give the details of the curve shapes.
In general adiabatic curves are steeper than isothermal curves...I realize this is not a mathematically rigorous statement but it is helpful ...I hope.
Ps...make sure you are clear about this before you worry about the details of monatomic, diatomic molecules etc !
Sir i know about that but i am not able to get mathemtical result, if possible could you please send me those.
Thanks
 
Chestermiller said:
Do you know the solution to that differential equation that you wrote down? If so, what is it?
Sir, i am not able to get the mathematical result
 
  • #10
Shreyans Jain said:
Sir, i am not able to get the mathematical result
Are you saying that you have not had a course in differential equations, and do not know how to solve a homogeneous first order linear ordinary differential equation?
 
  • #11
Chestermiller said:
Are you saying that you have not had a course in differential equations, and do not know how to solve a homogeneous first order linear ordinary differential equation?
I know how to solve a homogeneous first order linear ordinary differential equation but there was something missing which i have got it now now thanks.
 

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