Pay load 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 27°C . (Density of air = \( 1.2 kg m^{–3} \)And R = \( 0.083\; bar\; dm^3 K^{–1} mol^{-1}\) ).

Asked by Pragya Singh | 1 year ago | 68

Given:

r = 10 m

Therefore, volume of the balloon

= \( \dfrac{4}{3}πr^3\)

= \( \dfrac{ 4 }{ 3 }\; \times \; \dfrac{ 22 }{ 7 } \; \times \; 10^{3}\)

= 4190.5 m^{3} (approx.)

Therefore, the volume of the displaced air

= 4190.5 × 1.2 kg

= 5028.6 kg

Mass of helium,

= \( \dfrac{ MpV }{ RT }\)

Where, M = 4 × 10^{-3} kg mol^{-1}

p = 1.66 bar

V = volume of the balloon

= 4190.5 m^{3}

R =\( 0.083 0.083 \;bar\; dm^{ 3 } at \;K^{-1} mol^{-1}\)

T = 27°C = 300 K

Then,

\( m = \dfrac{ 4 \; \times \; 10^{-3} \; \times \; 1.66 \; \times \; 4190.5 \; \times \; 10^{3}}{0.083 \; \times \; 300}\)

= 1117.5 kg (approx.)

Now, total mass with helium,

= (100 + 1117.5) kg

= 1217.5 kg

Therefore, pay load,

= (5028.6 – 1217.5)

= 3811.1 kg

Therefore, the pay load of the balloon is 3811.1 kg.

Answered by Pragya Singh | 1 year agoExplain the physical significance of Van der Waals parameters.

Critical temperature for carbon dioxide and methane are 31.1°C and – 81.9°C respectively. Which of these has stronger intermolecular forces and why?

In terms of Charles’ law explain why -273°C is the lowest possible temperature.

What would be the SI units for the quantity \(\dfrac{pV^2 T^2}{n}\)

A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.