Does hot air require more work to compress than cold air?

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SUMMARY

Hot air requires more energy to compress than cold air due to the ideal gas law (PV=nRT). The work needed for compression can be calculated using the integral \(\int -P\,dV\). As temperature increases, the power required for compression at a constant pressure inlet and outlet also increases. This relationship is crucial for understanding the efficiency of physical systems involving gas compression.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Basic knowledge of thermodynamics and gas behavior
  • Familiarity with calculus, specifically integration
  • Concept of pressure-volume work in thermodynamic processes
NEXT STEPS
  • Study the implications of the ideal gas law on gas compression
  • Learn about thermodynamic cycles and their efficiency
  • Explore the relationship between temperature and pressure in gases
  • Investigate practical applications of gas compression in engineering systems
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Students and professionals in physics, engineering, and thermodynamics, particularly those interested in gas behavior and energy efficiency in compression systems.

Nabo00o
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Okey, simple question. As written in the title, I'm really not sure how this should work.
Should hot air need more energy to compress than cold air, is it the other way around or is there no difference?

This is asked because I need to find out if a physical system could work or not...
Any help appreciated!

Naboo
 
Engineering news on Phys.org
Hi Nabo00o, welcome to PF. One way to tackle this is to assume ideality (i.e., PV=nRT) and calculate the work needed to compress the gas, \int -P\,dV. Does this help?
 
Yes, the warmer the gas, the more power needed to compress given any pressure inlet and outlet and flow rate.
 
@Mapes
Thank you Mapes.
Yes I think that will help me as soon as I understand the formula correctly.

@Q_Goest
Okey. But is this only because air closer to total zero is harder to expand, or is it more or less proportional with the change of temperature?
Sorry if this is a dumb question, I just need to understand it correctly.
 

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