Archimedes tells us the lifting power of a balloon (how much mass it can lift) is equal to the difference between the mass of th

Question

3 years ago

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Archimedes tells us the lifting power of a balloon (how much mass it can lift) is equal to the difference between the mass of th

e balloon and the mass of the air it displaces. That is, if the balloon occupies the same volume as 10 kg of air, but the balloon only weighs 2 kg, then the balloon can lift 8 kg. gas densityH2 0.09 kg/m3He 0.18 kg/m3Air 1.23 kg/m3 Airships have sometimes been filled with hydrogen (H), but hydrogen is very flammable, and after the Hindenburg caught fire and crashed in 1937 with serious loss of life airships have nearly always been filled with helium (He) instead. Suppose a new airship Star of Balogna will have the shape of a cylinder 40. m long, with a radius of 4.0 m, and will weigh 750. kg without any gas inside it. Suppose also the average passenger weighs 80 kg. Calculate how many passengers the Balogna could carry if it were filled with hydrogen, and also if it were filled with helium. Passengers with Hz: 29Passengers with He: 26

Chemistry

1 answer:

Finger [1] · Answer 1 · 2022-09-04T23:00:20+03:00

Answer:

the Balogna could carry 19 passengers if filled with Hydrogen.

the Balogna could carry 17 passengers if filled with Helium.

Explanation:

Given that:

the radius of the cylinder r = 4.0 cm

the height h = 40 m

the volume of the cylinder will be:

$V = \pi r^2 h$

V = π (4.0 m)² (40)

V = 2010.62 m³.

where the density of the air = 1.23 kg/m³

the mass of the displaced air = Volume V ×density of the air

= 2010.62 m³ × 1.23 kg/m³

= 2473.0626 kg

To calculate how many passengers the Balogna could carry if it were filled with hydrogen, we have the following:

We are given the density of H₂ to be = 0.090 kg/m³

Therefore, the mass of Helium in the balloon = Volume V × density of the hydrogen

= 2010.62 m³ × 0.090 kg/m³

= 180.9558 kg

The total mass of the balloon = mass of H₂ + mass of empty balloon.

The total mass of the balloon = (180.9558+ 750) kg

= 930.9558 kg

Now, the Mass difference = mass of the air displaced -total mass of the balloon.

the Mass difference = (2473.0626 - 930.9558 ) kg

the Mass difference = 1542.1068 kg

∴

number of persons = mass difference/ mass of passenger

the number of persons = 1542.1068 kg /80 (kg/person)

the number of persons = 19.27 persons

the number of passengers $\simeq$ 19 passengers.

To calculate how many passengers the Balogna could carry if it were filled with helium, we have the following:

We are given the density of Helium to be = 0.18 kg/m³

Therefore, the mass of Helium in the balloon = Volume V × density of the hydrogen

= 2010.62 m³ × 0.18 kg/m³

= 361.9116 kg

The total mass of the balloon = mass of Helium + mass of empty balloon.

The total mass of the balloon = (361.9116 + 750) kg

= 1111.9116 kg

Now, the Mass difference = mass of the air displaced -total mass of the balloon.

the Mass difference = (2473.0626 - 1111.9116 ) kg

the Mass difference = 1361.151 kg

∴

number of persons = mass difference/ mass of passenger

the number of persons = 1361.151 kg /80 (kg/person)

the number of persons = 17.01 persons

the number of passengers $\simeq$ 17 passengers.