Pressure/Temp/Volume question

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SUMMARY

This discussion focuses on the isothermal expansion of one mole of an ideal gas at 300K, transitioning from a pressure of 2x10^6 N/m^2 to 2x10^5 N/m^2. The key equations involved include the ideal gas law (PV = nRT), the work done by the gas (W = PΔV), and the relationship between heat, work, and internal energy (ΔE = Q - W). The internal energy change (ΔE) is zero due to the isothermal condition, leading to the conclusion that heat absorbed (Q) equals work done (W). The challenge lies in determining the change in volume, which requires calculating the initial and final volumes using the ideal gas law.

PREREQUISITES
  • Understanding of the ideal gas law (PV = nRT)
  • Knowledge of thermodynamic principles, particularly isothermal processes
  • Familiarity with concepts of work and internal energy in thermodynamics
  • Ability to interpret PV diagrams for gas expansion
NEXT STEPS
  • Calculate the initial and final volumes using the ideal gas law (PV = nRT)
  • Explore the derivation of work done during isothermal expansion (W = ∫PdV)
  • Study the concept of entropy change for reversible processes in thermodynamics
  • Review the relationship between heat absorbed and work done in isothermal conditions
USEFUL FOR

Students studying thermodynamics, physics enthusiasts, and anyone seeking to understand the principles of gas behavior during isothermal expansion.

itr
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Homework Statement



One mole of an ideal gas at 300K expands isothermally from a pressure of 2x10^6 N/m^2 to a pressure of 2x10^5 N/m^2.

Derive the equations for, and calculate:

the heat absorbed by the gas...
the change in the internal energy of the gas...
the entropy change in the gas if it expands reversibly

Homework Equations



Q=mc(Δt)
ΔE = Q - W
PV = nRT
W = PΔV
Another equation involving ΔE, but i am not sure what it is...

The Attempt at a Solution



Since it is an isothermic equation, there is no Δt...which makes the ΔE equal zero(the internal energy). Since the internal energy is zero, that makes the equation ΔE=Q-W into Q=W. I am not sure how to finish the problem, because i have a change in pressure...but there is no change in temperature...
 
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Work is just the area under a PV diagram. Draw out the diagram, label V1 and V2 (the initial and final volumes), and you'll see what the area is.
 
how do i know what the change in volumes are? all i know is there is a pressure decrease, which could mean anything...unless i am missing something
 

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