Ideal Gases: dU=dW+dQ and dU=3/2RdT

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Discussion Overview

The discussion revolves around the relationships between the internal energy change (dU), work done (dW), and heat transfer (dQ) in the context of ideal gases. It explores the implications of two equations: dU = dW + dQ and dU = (3/2)RdT, particularly in scenarios involving changes in volume, pressure, and temperature.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions how dU in the general equation (I) can equal dU in the specific case (II) when compressing a gas, suggesting that changes in volume could affect pressure or temperature.
  • Another participant cautions against confusing the general equation of state (I) with the specific conditions of the solved equation (II), emphasizing the importance of understanding which terms go to zero under certain conditions.
  • A later reply expresses confusion about the classification of equation (II) as adiabatic, seeking clarification on the conditions under which the equations apply.
  • Another participant clarifies that (II) is not adiabatic but relates to conditions where volume is held constant (isochoric), referencing external sources for further context.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the equations and the conditions under which they apply. The discussion remains unresolved regarding the specific implications of the equations in various thermodynamic processes.

Contextual Notes

There are limitations in understanding which terms are considered zero in the general equation under specific conditions, and the discussion highlights the need for clarity on the assumptions made in each case.

georg gill
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(I):

dU=dW+dQ

also (II):

dU=\frac{3}{2}RdT

if you compress a gas dW in dU is positive from pV=nRT lesser volume could either mean more pressure or more T. If dV gives dp only then dT=0 how can then dU for dW in (I) be equal dU in (II)?
 
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George; don't confuse your general equation of state (I) with the solved equation of state with respect to the setting of a constant T, p, or V or adiabatic condition (II).

What is implied is the question of what term in the general equation goes to zero when the solved equation is presented for a process; then using the conditions that create that solution, allows one to use the ideal gas law to show relationships of the variables that are not being held constant.

Do you see where this is going now?
 
MrSid said:
George; don't confuse your general equation of state (I) with the solved equation of state with respect to the setting of a constant T, p, or V or adiabatic condition (II).

What is implied is the question of what term in the general equation goes to zero when the solved equation is presented for a process; then using the conditions that create that solution, allows one to use the ideal gas law to show relationships of the variables that are not being held constant.

Do you see where this is going now?

I am a bit confused unfortunately. Why is (II) adiabatic?

thank you
 
sorry George- not to imply that (II) is adiabatic;

but just to set the stage that one of those four conditions as constant when the equation of state is solved- In fact you will find in the wiki article Table_of_thermodynamic_equations
about halfway down an Equation Table for ideal gas and for the thermodynamic equation that matches (II) the variable that is kept constant is volume (isochoric).
 

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