A player bounces a basketball on the floor, compressing it to 80.5 % of its orig

Question

A player bounces a basketball on the floor, compressing it to 80.5 % of its original volume. The air (assume it is essentially N2 gas) inside the ball is originally at a temperature of 20.0C and a pressure of 2.00 atm. The ball's diameter is 23.5 cm . Part A What temperature does the air in the ball reach at its maximum compression? Assume that the compression of the air during the bounce is adiabatic. T = 46.6 C Correct Part B By how much does the internal energy of the air change between the ball's original state and its maximum compression? U = ___ J I got part A, and I see several answers for part B with the formula: T2=T1*(v1/v2)^y-1 and I cannot find the formula in my book and the professor never taught it. I cannot for the life of me figure out how to work out the math for this. Can someone please explain?! I want to understand it. Thanks!

Explanation / Answer

Start with the first law Q-W=U now no heat is being added to the system so Q=0 You're left with U= -W. What is the work done on the ball?
W=nC_v(T_1-T_2)
OR

•VF = 0.85 V0 = 5.7x10-3 m3

•Q constant à Adiabatic

•T0(V0)g-1 = TF(VF)g-1 , g=1.4

•TF = 320 K

•Ideal Gas: del U = CVnDT

• Air: CV ~ 20 J/(mol K)

del U = 345 J

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