Thermodynamis of air in a basketball

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SUMMARY

The discussion centers on the thermodynamics of air within a basketball during compression. The air, primarily nitrogen (N2), is initially at 20.0°C and 2.00 atm pressure, and upon compression to 80% of its original volume, the temperature rises to 47.4°C. The change in internal energy (ΔU) is calculated using the formula ΔU = nC_vΔT, where C_v is 20.76 J/(mol·K). The participant initially calculated ΔU as 0.00333657 J, but identified the need to apply the first law of thermodynamics, concluding that ΔU = -W, where W is the work done on the ball.

PREREQUISITES
  • Understanding of the ideal gas law and properties of gases
  • Familiarity with the first law of thermodynamics
  • Knowledge of specific heat capacities, particularly C_v
  • Basic algebra for manipulating thermodynamic equations
NEXT STEPS
  • Study the ideal gas law and its applications in thermodynamics
  • Learn about the first law of thermodynamics and its implications
  • Explore the concept of specific heat capacities, focusing on C_v for gases
  • Investigate the work done on gases during compression processes
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying thermodynamics, as well as engineers and anyone interested in the physical principles governing gas behavior in confined spaces.

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Homework Statement


A player bounces a basketball on the floor, compressing it to 80.0 % of its original volume. The air (assume it is essentially N_2 gas) inside the ball is originally at a temperature of 20.0 C and a pressure of 2.00 atm. The ball's diameter is 23.9 cm

Temperature at maximum compression equals 47.4 C

By how much does the internal energy of the air change between the ball's original state and its maximum compression?
\DeltaU=?

Homework Equations



\DeltaU=nC_v\DeltaT
C_v=20.76

The Attempt at a Solution



n=.0000058657

\DeltaU=.00333657
but that's not right
 
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Start with the first law

Q-W=ΔU

now no heat is being added to the system so Q=0

You're left with ΔU= -W. What is the work done on the ball?
 


W=nC_v(T_1-T_2)
that is the same thing i was using
 

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