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 temperature and atmospheric at sea level).

Explanation

The volume of spherical balloon of radius r,
`V=(4)/(3)pir^(3)`
At sea level, volume `V_(1)=(4)/(3)pir_(1)^(3)`
`(r_(1)" is the radius of balloon at sea level")`
`r_(1)=(0.2)/(2)m=0.1m`
`V_(1)=(4)/(3)pi(0.1m)^(3)`
Pressure at sea level, `p_(1)="1 atm"`
`"Volume at altitude, "V_(2)=(4)/(3)pir_(2)^(3)`
`(r_(2)" is the radius of balloon altitude")`
Pressure at the altitude = 0.65 atm.
According to Boyle.s law (since temperature is constant)
`p_(1)V_(1)=p_(2)V_(2)" or "V_(2)=(p_(1)V_(1))/(p_(2))`
`"or "(4)/(3)pir_(2)^(3)=("(1 atm)"xx(4)/(3)pi(0.1m)^(3))/("0.65 atm")`
`r_(2)^(3)=(1xx(0.1)^(3))/(0.65)`
`=1.54xx10^(-3)m^(3)`
`"or "r_(2)=(1.54xx10^(-3)m^(3))^(1//3)`
`=0.1154m`
`therefore" Diameter of the ballon at the altitude "=2xx0.1154`
`=0.2308m.`