A hot air balloon is a sphere of radius 8 m. The sir inside is at a temperature of 60^{\circ}C. How large a mass can the balloon lift when the outside temperature is 20^{\circ}C?

Answers (1)

The amount of pressure outside a balloon is lesser than the amount of pressure inside the balloon.

P_{i}-P_{0}=\frac{2\sigma}{R} {\sigma: surface tension, R: balloon’s radius}

Pi V = ni R Ti

Where, V is the volume of balloon and ni is number of moles of gas in the balloon

R is the gas constant, Ti is the temperature

ni = \frac{Pi V}{R Ti }= Mass of balloon (Mi)/Molecular mass (Ma)

Mi = \frac{Pi V Ma}{R Ti }-------------(1)

In the same way, n_{0} = \frac{P_{0} V}{R T_{0}}

According to principal of floatation, W + Mi g = M_{0} g, here W is the weight which the balloon lifts

W= (M_{0} - M_{i}) g

If the mass displaced by balloon is M0 and Ma is the mass inside or outside of the balloon

n_{0} = \frac{M_{0}}{M_{a}} = \frac{P_{0} V}{R T_{0}}

M_{0} =\frac{P_{0} V M_{a}}{R T_{0}}---------------(2)

Hence from (1) and (2) we can write,

W = \left [ \frac{P_{0} V Ma}{R T_{0} } -\frac{P_{i} V M_{a}}{R T_{i}}\right ] g

W = \frac{V M_{a}}{R}\left [ \frac{P_{0}}{T_{0}}-\frac{P_{i}}{T_{i}} \right ] g

Ma = 79% N2 + 21%O2

Ma = 0.79 \times 28 + 0.21 \times 32 = 28.84 g = 0.2884 kg

P_{i}-P_{0}=\frac{2\sigma}{R}

P_{i} = P_{o} + Pressure due to ST of membrane

Pi = \left [1.1013 \times 10^{5} + 2\times \frac{5}{8} \right ] = \sim 1.1013 \times 10^{5}

W = \frac{4}{3} \times \frac{3.14 \times 8 \times 8 \times 8\times 0.02884}{3\times 8.314} \left [\frac{1.013 \times 10^{5}}{293} - \frac{1.013 \times 10^{5}}{333} \right ]g

W = \frac{4}{3} \times \frac{3.14 \times 8 \times 8 \times 8\times 0.02884\times 1.013 \times 10^{5} }{3\times 8.314} \left [\frac{1}{293} - \frac{1}{333} \right ]g

W = 3044.2 N

 

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