Let the pressure inside the balloon be `p_(i)` and the outside pressure be `p_(@)`, then excess pressure is `p_(i)-p_(@)=(2S)/(r)` .
Where , S=Surface tension
r=radius of balloon
Considering the air to be in ideal gas `p_(i)V=n_(i)RT_(i)` where, V is the volume of the air displaced and `n_(@)` is the number of moles displaced and `T_(@)` is the temperature outside.
So, `n_(i)=(P_(i)V)/(RT_(i))=(M_(i))/(M_(A))`
Where, `M_(i)` is the mass of air inside and `M_(A)` is the molar mass of air
and `n_(@)=(P_(@)V)/(RT_(@))=(M_(@))/(M_(A))`
Where, `M_(@)` is the mass of air outside that has been displaced . If w is the load it can raise, then `w+M_(i)g=M_(@)g`
`implies w=M_(@)g-M_(i)g`
As in atmosphere 21% `O_(2)` and `79% N_(2)` -is present
`therefore` Molar mass of air
`M_(i)=0.21xx32+0.79xx28=28.84g`
`therefore` Weight raised by the balloon
`w=(M_(@)-M_(i))g`
`implies w=(M_(A)V)/(R)((P_(@))/(T_(@))-(P_(i))/(T_(i)))g`
`=(0.02884xx(4)/(3)pixx8^(3)xx9.8)/(8.314)((1.013xx10^(5))/(293)-(1.013xx10^(5))/(333)-(2xx5)/(8xx313))`
`=(0.02884(4)/(3)pixx8^(3))/(8.314)xx1.013xx10^(5)((1)/(293)-(1)/(333))xx9.8`
`=3044.2 N`
`therefore` Mass lifted by the balloon `=(w)/(g)=(3044.2)/(10)~~304.42` kg .
`~~305 kg`.
