Change in temprature is zero. what about change in internal energy?

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SUMMARY

The discussion centers on the relationship between temperature change and internal energy in thermodynamic processes. It establishes that while internal energy is typically a function of temperature, specific conditions can lead to changes in internal energy even when temperature remains constant. The example provided illustrates an isothermal expansion of gas under external pressure, where internal energy changes due to heat exchange with the surroundings. The conversation emphasizes the distinction between isothermal and adiabatic processes, highlighting that real gases may not conform to ideal gas behavior.

PREREQUISITES
  • Understanding of thermodynamic processes, specifically isothermal and adiabatic processes.
  • Familiarity with the ideal gas law and its limitations regarding real gases.
  • Knowledge of the first law of thermodynamics, particularly the relationship between internal energy, heat, and work.
  • Basic principles of heat transfer and energy conservation in closed systems.
NEXT STEPS
  • Study the first law of thermodynamics in detail, focusing on internal energy calculations.
  • Research Joule-Kelvin cooling and its implications for real gases.
  • Explore the differences between ideal and real gas behavior under various thermodynamic conditions.
  • Learn about heat transfer mechanisms in isothermal and adiabatic processes.
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Students and professionals in physics and engineering, particularly those studying thermodynamics, as well as anyone interested in the behavior of gases under varying pressure and temperature conditions.

vkash
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change in internal energy is a function of temperature. So if there is no change in temperature then there should no change in internal energy.
But consider this example.
A gas is at 20 atm pressure in a room whose pressure is 1 atm.using external forces(on piston). Gas slowly expands its volume till it's pressure became 1 atm.during the process the temperature is maintained at 278K(room temperature).
work done in process
work done(W)= nRT*ln(P1/P2) (Δ)T=0; so it's isothermal)
heat supplied(Q)=0 (No heat is supplied, gas expand by force on piston of it's own pressure)
change in internal energy=Q-W (using Q=U+W)
So from here we come to know that change in internal energy is not zero. But there is no change in temperature during the whole process so change in internal energy should zero...
So where am i getting it wrong.?

thanks!
 
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you will need to understand one of the basics in such contrived thermodynamic questions - isothermal (ie same temperature) and adiabatic (ie same energy) processes - in your example the internal energy changed because heat 'leaked out' into the room - overall energy was conserved
 
sambristol said:
you will need to understand one of the basics in such contrived thermodynamic questions - isothermal (ie same temperature) and adiabatic (ie same energy) processes - in your example the internal energy changed because heat 'leaked out' into the room - overall energy was conserved

I think you want to say that i am using both the processes in single case.(Q=0 as well as ΔT=0)
But what if we do this processes in a container having adiabatic walls(fully insulated walls). Will temperature of gas fall down?
 
yes - the kinetic energy of the escaped gas more than compensates for the specific heat capacity of the remaining gas* so it will cool - see footnote*

Regards

Sam

* Footnote This is for air at normal STP but all gases have an inversion temperature which is why we can liquidfy gases - look up Joule Kelvin Cooling
 
As they said, you can't consider there isn't heat exchange because in order to maintain the same temperature, you must exchange heat.

Also, internal energy is a function of temperature only if we are talking about ideal gases. This property does not hold for real gases.
 

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