Find entropy change for free expantion of ideal gas

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SUMMARY

The discussion focuses on calculating the entropy change (ΔS) for the free expansion of an ideal gas from a volume of 1 liter to 2 liters in an isolated system. Despite the energy and temperature remaining constant, the entropy does increase during this process. The formula for entropy change is defined as ΔS = ∫dS = ∫dQ/T, emphasizing the need for a reversible path to accurately determine the entropy change between the initial state (P, V, T) and the final state (P/2, 2V, T).

PREREQUISITES
  • Understanding of the ideal gas law
  • Familiarity with the concept of entropy in thermodynamics
  • Knowledge of reversible processes and their significance
  • Basic calculus for evaluating integrals
NEXT STEPS
  • Study the ideal gas law and its implications for thermodynamic processes
  • Learn about the calculation of entropy changes in various thermodynamic scenarios
  • Explore the concept of reversible and irreversible processes in thermodynamics
  • Investigate the relationship between work, heat flow, and entropy in ideal gases
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Students and professionals in thermodynamics, physicists, and engineers interested in understanding entropy changes in ideal gas expansions.

klinke
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entropy = jouls/kelvin
supose 1 liter of ideal gas is allowed to freely expand into a 2 liter volume in an isolated system
the energy in the system would remain the same,
the temperature in the system would remain the same
therefore if entropy =jouls/kelvin the entropy would remain the same
howerer if 1 liter of ideal gas is allowed to freely expand into a 2 liter volume in an isolated system the entropy does increase.
how dose entropy=jouls/kelvin ?
 
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klinke said:
entropy = jouls/kelvin
supose 1 liter of ideal gas is allowed to freely expand into a 2 liter volume in an isolated system
the energy in the system would remain the same,
the temperature in the system would remain the same
therefore if entropy =jouls/kelvin the entropy would remain the same
howerer if 1 liter of ideal gas is allowed to freely expand into a 2 liter volume in an isolated system the entropy does increase.
how dose entropy=jouls/kelvin ?
ΔS = ∫dS = ∫dQ/T over a reversible path between the initial and final states.

The initial state is (P,V,T) and the final state is (P/2,2V,T). So to calculate the entropy change you have to find a reversible path between those two states. (hint: the reversible path involves work being done and heat flow into the gas).

AM
 

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