Question

A balloon having a capacity of 10000 $m^3$ is filled with helium at ${20}^{o}C$ and 1 atm pressure. If the balloon is loaded with 80% of the load that it can lift at ground level, at what height will the balloon come to rest? Assume that the volume of the balloon is constant, the atmosphere is isothermal, ${20}^{o}C$, the molecular weight of air is 28.8 and the ground level, pressure is 1 atm. The mass of balloon is $1.3\times {10}^{6}g$.

• A. $1.4\times {10}^{-1}m$
• B. $2.4\times {10}^{3}m$
• C. $1.40\times {10}^{3}m$
• D. $2.4\times 10m$

Answer 1

Answer: (C)

$1.40\times {10}^{3}m$

At ground level $(Z=0)$
Atmospheric pressure $P=1atm$
Volume $V={10}^4{m}^3$ He
Temperature $T=20+273.15K$
At height h
Pressure $P=?$
Volume $V={10}^4{m}^3$ He
Temperature $T=20+273.15K$
Mass of empty ballon $=1.3\times {10}^6$ g
At equilibrium on the ground
Total mass $=m_{balloon}+m_{load}+m_{He}$
To achieve equilibrium at $Z=h$
Total mass $=m_{balloon}+0.80\left(m_{load}\right)+m_{He}$
According to Archimedes principle, the displaced air supports the weight of balloon + load.
At height $z=h$
The mass of displaced air $=m_{balloon}+0.80\left(m_{load}\right)+m_{He}$
At equilibrium on the ground
The mass of displaced air $=m_{balloon}+m_{load}+m_{He}$
Barometric formula
$\frac{P}{P_0}=\frac{\rho }{{\rho }_0}=ex{p}^{-Mgh/RT}$
Ideal gas behaviour
$PV=nRT$
Units $J=kg{m}^2{s}^{-2}$
Assumption: The volume of air displaced by the load is neglected in comparison to the volume of balloon.
At ground level, the mass of displaced air $={\rho }_{0}V$
At height h, the mass of displaced ait $=\rho V$
The equilibrium conditions are
$m_{balloon}+m_{load}+m_{He}={\rho }_{0}V$ ......(1)
$m_{balloon}+0.80\left(m_{load}\right)+m_{He}=\rho V$ ......(2)
Divide equation (2) by equation (1)
$\frac{\rho }{{\rho }_0}=\frac{m_{balloon}+0.80\left(m_{load}\right)+m_{He}}{m_{balloon}+m_{load}+m_{He}}$ ......(3)
$m_{He}=M_{He}\left(\frac{PV}{RT}\right)$
$m_{He}=\frac{4.0g/mol\times 1atm\times {10}^4{m}^3\times {10}^{3}L/m^3}{8.20578\times {10}^{-2}Latm/mol/K\times 293.15K}=1.663\times {10}^{6}g$
${\rho }_{0}V=m_{air}=\frac{28.8g/mol\times 1atm\times {10}^4{m}^3\times {10}^{3}L/m^3}{8.20578\times {10}^{-2}Latm/mol/K\times 293.15K}=1.1972\times {10}^{7}g$
From equation (1)
$m_{load}=1.1972\times {10}^{7}g-[1.3\times {10}^{6}g+1.663\times {10}^{6}g]=9.01\times {10}^{6}g$
Substituting in equation (3)
$\frac{\rho }{{\rho }_0}=\frac{1.3\times {10}^6+1.663\times {10}^6+0.80\times 9.01\times {10}^6}{1.3\times {10}^6+1.663\times {10}^6+9.01\times {10}^6}=0.85$
$\frac{\rho }{{\rho }_0}=0.85=ex{p}^{-\frac{Mgh}{RT}}$
$0.85=ex{p}^{-\frac{28.8g/mol\times 1kg/1000g\times 9.80665m/s^2\times hm}{8.31451J/K/mol\times 293.15K}}$
$0.85=ex{p}^{1.159\times {10}^{-4}h}$
$h=1.402\times {10}^{3}m$