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 1 of 4: Mass of air in the ball
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m=p Mair VRT=101325×28.82×1.131×10−48.314×298.15=0.133 gm = \dfrac{p\,M_{\text{air}}\,V}{RT} = \dfrac{101325\times 28.82\times 1.131\times 10^{-4}}{8.314\times 298.15} = 0.133\ \text{g}
Analysis

V=43π(3.0 cm)3=113.1V = \tfrac{4}{3}\pi(3.0\text{ cm})^3 = 113.1 cm3^3 and Mair=0.8×28.02+0.2×32.00=28.82M_{\text{air}} = 0.8\times 28.02 + 0.2\times 32.00 = 28.82 g mol−1^{-1}, so m=pMairV/RT=0.133m = pM_{\text{air}}V/RT = 0.133 g.