A player bounces a basketball on the floor, compressing it to 80.0 % of its original volume. The air (assume it is essentially N2 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. By how much does the internal energy of the air change between the ball's original state and the maximum compression?

We are given the initial conditions of the ball:

T1 = 20 degrees Celcius

P1 = 2 atm

d1 = 23.9 cm = 0.239 m

V1 = 4/3 * pi * r^3

V1 = 4/3 * 3.14 * (0.239 m) ^3

V1 = 0.057 m^3

V2 = 0.8V1

V2 = 0.0457 m^3

We are to determine the change in internal energy of the air in the ball:

ΔU = Q + W

where Q is zero since there is no heat transfer

ΔU = W = ΔPV where P is constant

ΔU = PΔV

ΔU = 2 atm * 101325 Pa * (0.0457 - 0.0571) m^3

ΔU = - 2310.21 Joules