Final Temperature and Pressure Increase in Adiabatic Air Compression?

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SUMMARY

The final temperature of air compressed adiabatically from an initial temperature of 20°C (293.15 K) by a factor of 15 is calculated to be approximately 866.017 K. The pressure increases by a factor of 44.31 during this compression process, as derived from the equations PVγ = constant and TVγ-1 = constant, with γ (gamma) for air set at 1.4. The calculations confirm the accuracy of these results, aligning with the ideal gas law for further validation.

PREREQUISITES
  • Understanding of adiabatic processes in thermodynamics
  • Familiarity with the ideal gas law
  • Knowledge of the specific heat ratio (γ) for air
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the adiabatic process equations
  • Learn about the implications of the ideal gas law in thermodynamic systems
  • Explore the concept of specific heat capacities and their role in thermodynamics
  • Investigate real-world applications of adiabatic compression in engineering
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Students in thermodynamics, engineers working with pneumatic systems, and anyone interested in the principles of adiabatic processes and gas laws.

rg2004
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I'd just like to check my answer, it seems too big

Homework Statement


Air initially at 20◦C is compressed by a factor of 15.

What is the final temperature assuming that the compression is adiabatic and \gamma ≈ 1.4 the value of \gamma for air in the relevant range of temperatures? By what factor does the pressure increase?

Homework Equations



PV\gamma=constant
TV\gamma-1=constant

The Attempt at a Solution



V1/15=V2
T1V1\gamma-1=T2V2\gamma-1
293.15*V1\gamma-1=T2*V1/15
293.15*15\gamma-1=T2
293.15*15.4=T2
T2=866.017 kelvinP1V1\gamma=P2V2\gamma
P1V1\gamma=P2V1/15\gamma
P1*15\gamma=P2
thus P2 increased by a factor of 44.31does it look right?
 
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Looks right. You can also use the ideal gas law to find final pressure:

nR = P_1V_1/T_1 = P_2V_2/T_2

P_2/P_1 = T_2V_1/T_1V_2 = \frac{866*15}{293} = 44.3

AM
 
Thanks!
 

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