Problem 4

Tennis balls are pressurised above atmospheric pressure for a good bounce; old hollow balls simply contained air at atmospheric pressure. Assume a ball inner radius

R = 3.0

cm constant during pressurisation, air composed of 20 % O2 and 80 % N2 by volume, and that overpressure does not expand the ball. (a) Calculate the mass

m

of air inside an old ball at

T_0 = 25

°C. (b) Modern balls are kept at

p_0 = 1.80

atm; premium balls contain pure N2. Once the can is opened, gas diffuses out until the inside reaches atmospheric pressure, with no inward diffusion; the process follows first-order kinetics in the overpressure. Premium balls depressurise to

p_1 = 1.40

atm after

t_1 = 241

h and to

p_2 = 1.19

atm after about

t_2 = 21

days at 25 °C. Demonstrate first-order behaviour and calculate the depressurisation rate constant

k_{\mathrm{N_2}}

in h

^{-1}

. (c) The initial depressurisation rate of regular (air) balls is 10 % faster than that of premium balls; calculate the rate constant

k_{\mathrm{O_2}}

assuming the rates of both gases are additive. (d) New premium balls, manufactured and canned at 25 °C, are opened and quickly brought to

T = 30.0

°C; a match starts

t = 12.0

h later. With

E_a = 50.0

kJ mol

^{-1}

for premium-ball depressurisation, calculate the ball pressure at the start of the match.

Step 2 of 4: First-order rate constant

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k_{\mathrm{N_2}} = \dfrac{1}{241}\ln\dfrac{0.80}{0.40} = 2.88\times 10^{-3}\ \text{h}^{-1};\ \dfrac{1}{504}\ln\dfrac{0.80}{0.19} = 2.85\times 10^{-3}\ \text{h}^{-1}

Analysis

Only the overpressure $\Delta p = p - 1.00$ atm decays: $\Delta p = 0.80, 0.40, 0.19$ atm at $t = 0, 241, 504$ h. Both intervals give $k = (1/t)\ln(\Delta p_0/\Delta p) \approx 2.86\times 10^{-3}$ h $^{-1}$ — the constancy proves first order; the accurate value uses the exact data point: $k_{\mathrm{N_2}} = 2.88\times 10^{-3}$ h $^{-1}$ .

Common pitfall. The decay variable is the overpressure p − 1 atm, not the total pressure; using total pressure gives inconsistent k values.