Thermodynamis of air in a basketball

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The discussion centers on a physics problem involving the thermodynamics of air in a basketball during compression. The basketball's air, initially at 20.0°C and 2.00 atm, is compressed to 80% of its volume, resulting in a maximum temperature of 47.4°C. The change in internal energy (ΔU) is calculated using the formula ΔU=nC_vΔT, with C_v given as 20.76. An attempt to solve for ΔU resulted in a value of 0.00333657, which was deemed incorrect, leading to a reevaluation using the first law of thermodynamics, where Q=0 and ΔU is equal to -W. The work done on the ball is expressed as W=nC_v(T_1-T_2), indicating the need for accurate calculations of n and temperature changes to determine the correct internal energy change.
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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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