Isothermal Expansion, No Calculus

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SUMMARY

The discussion focuses on calculating work done on a gas during isothermal expansion without using calculus. The gas starts at a pressure of 2.52e5 Pa in a container with a volume of 444 cm³ and expands until it reaches atmospheric pressure of 0.857e5 Pa. The work done on the gas can be calculated using the formula W = PVln(V2/V1), where V2 is the final volume and V1 is the initial volume. The change in internal energy is zero, and the heat exchanged equals the work done, with appropriate sign conventions applied.

PREREQUISITES
  • Understanding of the first law of thermodynamics
  • Familiarity with ideal gas laws
  • Basic knowledge of isothermal processes
  • Ability to manipulate logarithmic functions
NEXT STEPS
  • Research the derivation of the work formula for isothermal expansion
  • Study the implications of the first law of thermodynamics in thermodynamic processes
  • Explore the concept of internal energy in ideal gases
  • Learn about the relationship between pressure, volume, and temperature in gas laws
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, particularly those tackling problems related to gas laws and isothermal processes without calculus. It is also useful for educators seeking to explain these concepts in a simplified manner.

Alexandra_H
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Homework Statement



"We have some gas in a container at high pressure. The volume of the container is 444 cm^3. The pressure of the gas is 2.52e5 Pa. We allow the gas to expand at constant temperature until its pressure equals atmospheric pressure, which is 0.857e5 Pa."

A. Find the work (J) done on the gas.
B. Find the change of internal energy (J) of the gas.
C. Find the amount of heat (J) done on the gas.

Homework Equations



First law of thermodynamics. Ideal gas laws.

The Attempt at a Solution



(A) is what I'm having trouble with. The class is not calculus-based. (B) is zero, and for (C), W = Q (appropriate signs, - and -, as they seem to be referencing the surroundings, not the gas itself)--easy enough, once I have A.

Thoughts?
Thanks!
 
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Since you can't use calculus to derive the formula, I think you'd just need to use the direct result:

W=PVln(V2/V1)
 

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