Temperature Changes with Volume Increase: Charles Law

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SUMMARY

Charles Law states that a gas expands and increases its volume when heated, but the temperature change depends on the system's conditions. Specifically, in isothermal processes where internal energy remains constant, the equation pV=nRT applies. In isobaric processes, where pressure is constant and volume increases, the gas does work on its surroundings, leading to a temperature decrease if no heat is absorbed. Conversely, when external pressure is zero, the gas expands without doing work, maintaining a constant temperature.

PREREQUISITES
  • Understanding of Charles Law and its implications on gas behavior
  • Familiarity with the ideal gas law (pV=nRT)
  • Knowledge of isothermal and isobaric processes
  • Basic concepts of internal energy and thermodynamic work
NEXT STEPS
  • Study the implications of the ideal gas law in various thermodynamic processes
  • Explore the relationship between temperature, pressure, and volume in gases
  • Learn about the concept of internal energy in thermodynamics
  • Investigate real gas behavior versus ideal gas assumptions
USEFUL FOR

Students of physics, chemists, and anyone interested in thermodynamics and gas laws will benefit from this discussion.

Gersty
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Charles Law says that a gas will expand (increase its volume) as it is heated. Is it also true therefore that the temperature of a gas will rise if it's volume increases somehow?
 
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It depends on the system.

Charles' law is a spesific case of the ideal gas law,

[tex]pV=nRT[/tex]

In this equation, internal energy is held constant, meaning it applies for isothermal processes. In Charles' law the pressure and amount of gas is kept constant.

The following discussion is valid when the gas expands against a uniform external pressure.

If you have an isobaric process (constant pressure) where the volume increases, it means that the gas is doing work on the surroundings. In an ideal gas, there is no potential energy between particles, so

[tex]U= \sum E_k + 0=\frac{3}{2}nRT[/tex]

for a monoatomic gas. From there we get that

[tex]\Delta U=\frac{3}{2}nR\Delta T=q-w[/tex]

If there is no heat from the surroundings, the temperature will sink.

In the special case when the external pressure is zero, the gas expands without doing work and the temperature is constant.
 

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