Chemistry Labs

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=3.0R = 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 mm of air inside an old ball at T0=25T_0 = 25 °C. (b) Modern balls are kept at p0=1.80p_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 p1=1.40p_1 = 1.40 atm after t1=241t_1 = 241 h and to p2=1.19p_2 = 1.19 atm after about t2=21t_2 = 21 days at 25 °C. Demonstrate first-order behaviour and calculate the depressurisation rate constant kN2k_{\mathrm{N_2}} in h−1^{-1}. (c) The initial depressurisation rate of regular (air) balls is 10 % faster than that of premium balls; calculate the rate constant kO2k_{\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=30.0T = 30.0 °C; a match starts t=12.0t = 12.0 h later. With Ea=50.0E_a = 50.0 kJ mol−1^{-1} for premium-ball depressurisation, calculate the ball pressure at the start of the match.
Step 4 of 4: Pressure at the match
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pstart=1+(1.80303.15298.15−1)e−k(303.15) t=1.74 atmp_{\text{start}} = 1 + \left(1.80\dfrac{303.15}{298.15} - 1\right)e^{-k(303.15)\,t} = 1.74\ \text{atm}
Analysis

Three corrections combine: (i) warming to 303.15 K raises the initial pressure to 1.80×303.15/298.15=1.831.80\times 303.15/298.15 = 1.83 atm; (ii) Arrhenius raises the rate constant to k(303.15)=2.88×10−3exp⁡[500008.314(1298.15−1303.15)]≈4.0×10−3k(303.15) = 2.88\times 10^{-3}\exp[\frac{50000}{8.314}(\tfrac{1}{298.15}-\tfrac{1}{303.15})] \approx 4.0\times 10^{-3} h−1^{-1}; (iii) after 12 h the overpressure decays: p=1+0.83 e−4.0×10−3×12≈1.74p = 1 + 0.83\,e^{-4.0\times 10^{-3}\times 12} \approx 1.74 atm.