Question

In a test of the elasticity of basketballs, a basketball fired by an air gun against a solid wall, compressing it to 80.0 % of its original volume. The air inside the ball (assume it is all N2 gas) is originally at a temperature of 20.0 ∘C and a pressure of 2.00 atm. The ball's radius is 11.95 cm.

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.

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

Answer 1

Part A.

Given that process is adiabatic, So

PV^ = constant, OR

TV^( - 1) = constant, So

T2 = T1*(V1/V2)^( - 1)

= Cp/Cv = 1.40 for diatomic gases (N2)

V1 = V, then V2 = 0.80*V

T1 = 20 C = 293 K

So,

T2 = 293*(V/(0.80*V))^(1.40 - 1)

T2 = 293/0.80^0.40

T2 = 320.4 K = 47.4 C

Part B.

We know that Original volume of ball is given by:

V = 4*pi*R^3/3

Using ideal gas law:

PV = nRT

n = PV/RT

P = 2.00 atm = 2*1.01325*10^5 Pa

initially moles of air inside basketball will be:

n = 2.00*1.01325*10^5*(4*pi*(11.95*10^-2)^3/3)/(8.314*293)

n = 0.595 moles

Now change in internal energy after compression will be:

dU = n*Cv*dT

Cv for air = 20.79 J/mol.K (Please check this value in your reference book)

dT = 47.4 - 20 = 27.4

dU = 0.595*20.79*27.4

dU = 338.94 J

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