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 - mg = 0
where M is the mass of the balloon and its occupants, m is the mass of the hot gas inside the balloon.
the force acting upward F = Vρg
so
Vρg - Mg - mg = 0
solve for m
m = ( Vρg - Mg ) / g
m = Vρg/g - Mg/g
m = ρV - M ------- let this be equation 1
Now, from the ideal gas law, PV = nRT
we know that number of moles n = m / μ
where μ is the molecular mass of air
so
PV = (m/μ)RT
solve for T
μPV = mRT
T = μPV / mR -------- 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