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Payload is Defined as the Difference Between the Mass of Displaced Air and the Mass of the Balloon. Calculate the Payload When a Balloon of Radius 10 M, Mass 100 Kg is Filled with Helium at 1.66 Bar at 27°C - Chemistry

Payload is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload 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 m3 and R = 0.083 bar dm3 K1 mol1).

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Solution 1

Given,

Radius of the balloon, r = 10 m

∴ Volume of the balloon  = `4/3pir^3`

`= 4/3 xx22/7 xx 10^3`

`= 4190.5 m^3 ("approx")`

Thus, the volume of the displaced air is 4190.5 m3.

Given,

Density of air = 1.2 kg m–3

Then, mass of displaced air = 4190.5 × 1.2 kg

= 5028.6 kg

Now, mass of helium (m) inside the balloon is given by,

`m = (MpV)/(RT)`

Here

`M = 4xx 10^(-3) kg mol^(-1)`

p = 1.66  bar

V = Volume of the balloon

= `4190.5 m^3`

`R = 0.083 " bar" dm^3 K^(-1) mol^(-1)`

`T = 27 ^@C = 300 K`

Then `m = (4xx10^(-3)xx1.66xx4190.5xx10^3)/(0.083 xx300)`

= 1117.5 kg (approx)

Now, total mass of the balloon filled with helium = (100 + 1117.5) kg

= 1217.5 kg

Hence, pay load = (5028.6 – 1217.5) kg

= 3811.1 kg

Hence, the pay load of the balloon is 3811.1 kg

 

Solution 2

Radius of the balloon = 10 m

;.Volume of the balloon = `4/3 pi r^3` =  `4/3 xx 22/7 xx (10 m)^3  = 4190.5 m^3`

Volume of He filled at  `1.66 "bar" and 27^@C = 4190.5 m^3`

Calculation of mass of He

`Pv = nRT = omega/M RT` or `omega = (MPV)/(RT) =( (4xx10^(-3) kg mol^(-1))(1.66 "bar")(4190.5 xx 10^3 dm^3))/((0.083 "bar" dm^3 K^(-1) mol^(-1))(300K))`

=1117.5 kg

Total mass of the balloon along with He = 100 + 1117.5 = 1217.5 kg

Maximum mass of the air that can be displaced by balloon to go up =Volume xx Density

`= 4190.5 m^3 xx 1.2 kg m^(-3) = 5028.6 kg`

:. Payload = 5028.6 - 1217.5 kg = 3811.1 kg

Concept: The Gaseous State
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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 5 States of Matter
Q 16 | Page 153
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