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Gelneren [198K]

3 years ago

5

The mass of a hot-air balloon and its occupants is 381 kg (excluding the hot air inside the balloon). The air outside the balloo

n has a pressure of 1.01 x 105 Pa and a density of 1.29 kg/m3. To lift off, the air inside the balloon is heated. The volume of the heated balloon is 480 m3. The pressure of the heated air remains the same as that of the outside air. To what temperature in kelvins must the air be heated so that the balloon just lifts off

1 answer:

Kamila [148]3 years ago

5 0

Answer:

In order to lift off the ground, the air in the balloon must be heated to 710.26 K

Explanation:

Given the data in the question;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

let F represent the force acting upward.

Now in a condition where the hot air balloon is just about to take off;

F - Mg - m $_g$ g = 0

where M is the mass of the balloon and its occupants, m $_g$ is the mass of the hot gas inside the balloon.

the force acting upward F = Vρg

so

Vρg - Mg - m $_g$ g = 0

solve for m $_g$

m $_g$ = ( Vρg - Mg ) / g

m $_g$ = Vρg/g - Mg/g

m $_g$ = ρV - M ------- let this be equation 1

Now, from the ideal gas law, PV = nRT

we know that number of moles n = m $_g$ / μ

where μ is the molecular mass of air

so

PV = (m $_g$ /μ)RT

solve for T

μPV = m $_g$ RT

T = μPV / m $_g$ R -------- let this be equation 2

from equation 1 and 2

T = μPV / (ρV - M)R

so we substitute in our values;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]

T = 1405920 / 1979.442

T = 710.26 K

Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K

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3 years ago

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