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Exercises · 5.16

Q.Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m–3 and R = 0.083 bar dm3 K–1 mol–1).

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Step 1 – Volume of the balloon (sphere, radius 10 m)

V=43πr3=43π(10 m)3=43π(1000 m3)=4188.79 m3V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (10\ \text{m})^3 = \frac{4}{3}\pi (1000\ \text{m}^3) = 4188.79\ \text{m}^3

Step 2 – Mass of air displaced

mair displaced=V×dair=4188.79 m3×1.2 kg m−3=5026.55 kgm_{\text{air displaced}} = V \times d_{\text{air}} = 4188.79\ \text{m}^3 \times 1.2\ \text{kg m}^{-3} = 5026.55\ \text{kg}

Step 3 – Moles of He inside the balloon (ideal gas equation)

Convert VV to dm3^3 (1 m3=1000 dm31\ \text{m}^3 = 1000\ \text{dm}^3): V=4,188,790 dm3V = 4{,}188{,}790\ \text{dm}^3

T=27+273=300 K,p=1.66 bar,R=0.083 bar dm3K−1mol−1T = 27+273 = 300\ \text{K}, \qquad p = 1.66\ \text{bar}, \qquad R = 0.083\ \text{bar dm}^3\text{K}^{-1}\text{mol}^{-1}

n(He)=pVRT=(1.66)(4,188,790)(0.083)(300)=6,953,391.424.9=279,253.9 moln(\text{He}) = \frac{pV}{RT} = \frac{(1.66)(4{,}188{,}790)}{(0.083)(300)} = \frac{6{,}953{,}391.4}{24.9} = 279{,}253.9\ \text{mol}

Step 4 – Mass of He …

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