Work and Heat in Thermodynamic Processes

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SUMMARY

The discussion focuses on calculating work done by a gas during thermodynamic processes involving constant pressure. The gas initially at a pressure of 1.20 atm and a volume of 8.00 m³ expands to twice its volume, resulting in work calculated using the formula W = P(Vf - Vi). The pressure must be converted from atmospheres to Pascals for accurate calculations, as 1 atm equals 101325 Pa. Additionally, the work done during compression to one quarter of the initial volume is also addressed.

PREREQUISITES
  • Understanding of the ideal gas law
  • Familiarity with the concept of work in thermodynamics
  • Knowledge of unit conversions, specifically from atmospheres to Pascals
  • Proficiency in using the formula W = P(Vf - Vi)
NEXT STEPS
  • Learn about the ideal gas law and its applications
  • Study the principles of thermodynamic processes and work calculations
  • Explore unit conversion techniques, particularly pressure units
  • Investigate examples of work done in isothermal and adiabatic processes
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Students and professionals in physics and engineering, particularly those focusing on thermodynamics and gas laws.

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Gas in a container is at a pressure of 1.20 atm and a volume of 8.00 m3.

(a) What is the work done by the gas if it expands at constant pressure to twice its initial volume?
wrong check mark J
(b) If it is compressed at constant pressure to one quarter of its initial volume?

I am using the equation W=P(Vf-Vi)

The change in volume would 8.00 m3?
The pressure is in atm and I'm not sure what it should be converted to...

is any of this correct?
 
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Convert atmospheres to Pascals, the standard SI unit of pressure. (1 Pa = 1 N/m^2) Then your work will be in Joules.
 

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