A perfectly elastic spherical balloon of 0.02 m diameter was filled with hydrogen at sea level. What will be its diameter when it has risen to an altitude where the pressure is 0.65 atm ? (Assume no change in temjperature and atmospheric at sea level).

Explanation

If `r_(1)` is the radius of the balloon at sea level, then
volume of the balloon at sea level `=(4)/(3)pi r_(1)^(3)=(4)/(3)pi(0.1m)^(3)`
i.e.., volume of the gas at sea level `(V_(1))=(4)/(3)pi(0.1 m)^(3)`
Pressure at the sea level `(P_(1))=1 atm`
Suppose the radius of the balloon at altitude=`r_(2)`.Then
Volume of the balloon at altitude `(V_(2))=(4)/(3)pir_(2)^(3)`
Pressure at the altitude `(P_(2))=0.65 atm` (Given)
As temperature remains constant, applying Boyle's law
`{:(P_(1)V_(1),=,P_(2)V_(2)),(("At sea level"),,("At altitude")):}`
`1 atmxx(4)/(3)pi(0.1 m)^(3)=0.65 atmxx(4)/(3)pir_(2)^(3)" or " r_(2)^(3)=((0.1 m)^(3))/(0.65)=1.54xx10^(-3)m^(3)`
`:. r_(2)=(1.54xx10^(-3))^(1//3)m=0.1154" m"` ,
`:.` Diameter of the balloon at altitude`=2xx0.1154 m=0.2308 m`.