Ncert Exercises & Solutions

Q. 23 A hot air balloon is a sphere of radius 8 m. The air inside is at a temperature of 60°C. How large a mass can the balloon lift when the outside temperature is 20°C? Assume air is an ideal gas, R=8.314 J mol $e^{- 1} K^{- 1},$ 1 atm = 1.013 $\times 10^{5}$ Pa, the membrane tension is 5 $N m^{- 1}$ .

Hint: Apply the concept of excess pressure.

Step 1: Find the excess pressure inside the balloon.

Let the pressure inside the balloon be

$p_{i}$ and the outside pressure be

$p_{o}$ , then excess

pressure is

$p_{i} - p_{o} = \frac{2 s}{r}$

where S = Surface tension
r= radius of the balloon

Considering the air to be an ideal gas,

$p_{i} V = n_{i} R T_{i}$ where, V is the volume of the air inside the balloon,

$n_{i}$ is the number of moles inside the balloon and

$T_{i}$ is the temperature inside the balloon, and where V is the volume of the air displaced and

$n_{o}$ is the number of moles displaced and

$T_{o}$ is the temperature outside.

So.

$n_{i} = \frac{p_{i} V}{R T_{i}} = \frac{M_{i}}{M_{A}}$

where

$M_{i}$ is the mass of air inside and

$M_{A}$ is the molar mass of air.

and

$n_{o} = \frac{p_{o} V}{R T_{o}} = \frac{M_{o}}{M_{A}}$

where

$M_{o}$ is the mass of air outside that has been displaced.

Step 2: Find the weight lifted by the balloon.

If w is the load it can raise, then,

w +

$M_{i} g = M_{o} g$

$\Rightarrow w = M_{o} g - M_{i} g$

As in atmosphere

$21 % O_{2} a n d 79 % N_{2}$ are present.

$∴$ The molar mass of air,

$M_{i} = 0.21 \times 32 + 0.79 \times 28 = 28.84 g .$

$∴$ Weight raised by the balloon,

$w = (M_{o} - M_{i}) g \Rightarrow w = \frac{M_{A} V}{R} (\frac{p_{o}}{T_{o}} - \frac{p_{i}}{T_{i}}) g = \frac{0.02884 \times \frac{4}{3} π \times 8^{3} \times 9.8}{8.314} (\frac{1.013 \times 10^{5}}{293} - \frac{1}{333} (1.013 \times 10^{5} - \frac{2 \times 5}{8 \times 313})) = \frac{0.02884 \times \frac{4}{3} π \times 8^{3} \times 9.8}{8.314} \times 1.013 \times 10^{5} (\frac{1}{293} - \frac{1}{333}) = 3044.2 N$

$∴$ Masslifted by the balloon,

$m = \frac{w}{g} = \frac{3044.2}{10} \approx 304.42 k g \approx 305 k g .$

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