Answer:
Excess pressure
inside balloon,
\[{{P}_{i}}-\,{{P}_{o}}\,\,\frac{2\sigma
}{r}\] ... (i)
Here
\[\sigma \,=\] surface
tension
For
ideal gas, \[PV=\mu \,RT\]
\[\therefore
\] \[\mu =\,\frac{PV}{R{{T}_{i}}}\]
\[\text{=}\,\frac{\text{Mass}\,\,\text{of}\,\,\text{air}\,\,\text{inside}\,\,\text{balloon}\,\,\text{(}{{\text{M}}_{\text{i}}}\text{)}}{\text{Molar}\,\,\text{mass}\,\,\text{of}\,\,\text{air}\,\text{(}{{\text{M}}_{\text{A}}}\text{)}}\]
are
the number of moles inside balloon Similarly, number of moles outside the
balloon
\[{{\mu
}_{0}}=\,\frac{{{P}_{0}}V}{R{{T}_{0}}}\]
\[\text{=}\,\frac{\text{mass}\,\,\text{of}\,\,\text{air}\,\,\text{outside}\,\,\text{balloon}\,\text{(}{{\text{M}}_{\text{0}}}\text{)}}{\text{molar}\,\,\text{mass}\,\text{of}\,\text{air}\,\,\text{(}{{\text{M}}_{\text{A}}}\text{)}}\]
Molar
mass of air, \[{{M}_{A}}=28.84\,g\]
\[=28.84\times
{{10}^{-3}}\,kg\]
Weight
raised by balloon \[=({{M}_{o}}-{{M}_{i}})g\] ?. (ii)
Now
\[{{M}_{o}}=\frac{{{P}_{o}}V{{M}_{A}}}{R{{T}_{o}}}\] and \[{{M}_{i}}=\frac{{{P}_{i}}V\,{{M}_{A}}}{R{{T}_{i}}}\]
\[\therefore
\] Weight
raised, \[W=\,\frac{V{{M}_{A}}}{R}\,\left( \frac{{{P}_{0}}}{{{T}_{0}}}-\
\frac{{{P}_{i}}}{{{T}_{i}}} \right)g\,\] ...(iii)
From
eqn. (i), \[{{P}_{i}}=\,{{P}_{0}}+\,\frac{2\sigma }{r}\]
\[\therefore
\]\[W=\,\frac{V{{M}_{A}}}{R}\,\,\left[
\frac{{{P}_{o}}}{{{T}_{o}}}-\frac{{{P}_{o}}}{{{T}_{i}}}\,-\frac{2\sigma
}{r{{T}_{i}}} \right]\,g\]
or
mass raised, m
\[=\,\frac{V{{M}_{A}}}{R}\,\left[
{{P}_{o}}\,\left( \frac{1}{{{T}_{o}}}-\frac{1}{{{T}_{i}}} \right)-\frac{2\sigma
}{r{{T}_{i}}} \right]\]
\[=\,\frac{\frac{4}{3}\,\pi
{{r}^{3}}\,{{M}_{A}}}{R}\,\left[ {{P}_{o}}\left(
\frac{1}{{{T}_{o}}}-\frac{1}{{{T}_{o}}} \right)-\frac{2\sigma }{r{{T}_{i}}}
\right]\]
or \[m=\,\frac{4\times
\,3.14\,\times \,{{(8)}^{3}}\times \,28.84\,\times \,{{10}^{-3}}}{3\times
\,8.314}\]
\[\left[
1.013\times \,{{10}^{5}}\left( \frac{1}{293}-\frac{1}{373}
\right)\,-\frac{2\times \,5}{8\times \,333} \right]\]
\[=304.42\,kg\]
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