How Does Gas Behavior on a PV Diagram Indicate Changes in Internal Energy?

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SUMMARY

The discussion centers on the behavior of gas on a p-V diagram and its implications for internal energy changes. Specifically, when a gas moves along an isotherm, such as in processes (a) and (b), the change in internal energy (ΔEint) is zero, as internal energy is dependent on temperature changes. Conversely, moving into region 1 results in a negative ΔEint due to a decrease in temperature, while moving into region 2 yields a positive ΔEint due to an increase in temperature. This analysis is valid under the assumption of ideal gas behavior.

PREREQUISITES
  • Understanding of p-V diagrams and isotherms
  • Familiarity with the ideal gas law
  • Knowledge of internal energy concepts in thermodynamics
  • Basic calculus for interpreting changes in state variables
NEXT STEPS
  • Study the ideal gas law and its applications in thermodynamics
  • Learn about the first law of thermodynamics and its relation to internal energy
  • Explore the concept of isothermal processes and their characteristics
  • Investigate the implications of real gas behavior compared to ideal gases
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This discussion is beneficial for students studying thermodynamics, physics educators, and anyone interested in the principles of gas behavior and internal energy changes in ideal gases.

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Homework Statement

figure: http://www.chegg.com/homework-help/questions-and-answers/dot-figure-represents-initial-state-gas-isotherm-dot-divides-p-v-diagram-regions-1-2-follo-q5905441

The dot in Figure (b) represents the initial state of a gas, and the isotherm through the dot divides the p-V diagram into regions 1 and 2. For the following processes, determine whether the change ΔEint in the internal energy of the gas is positive, negative, or zero.

(a) the gas moves up along the isotherm(b) the gas moves down along the isotherm(c) the gas moves to anywhere in region 1(d) the gas moves to anywhere in region 21.ΔEint is positive
2.ΔEint is negative
3.ΔEint is zero

Homework Equations



curve of an isotherm is a hyperbola so as the hyperbola becomes closer to the center (0,0) at its inflection point the temperature decreases

deltaE internal= 3/2 nRdelta T means as temperature increases deltaE positive and Vise Versa

The Attempt at a Solution



when moving on isotherm the delta Einternal will be zero since Einternal depends on the change of temperature the a and b will be zero (a,b =3)

but while moving under the curve then we are getting a lesser temperature. the temperature decreases so delta E negative so c is negative (c= 2

similarly for moving above the curve then d is positive (d=1)

Is my answer correct?
 
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Yes, provided in a and b you are dealing with an ideal gas.
 
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