At sea level a balloon has volume of `785xx10^(-3) dm^(3)`. What will be its volume, if it taken to a place where the pressure is 0.052 atm. Less than the atmospheric pressure of 1 atm.

Explanation

At sea level pressure `P_(1)=1 atm`
Volume occupied at sea level `V_(1)=758xx10^(-3) dm^(-3)`
If the pressure `P_(2)=0.052 atm`
the valume of the balloon `V_(2)= ?`
According to Boyle's law
`P_(1)V_(1)=P_(2)V_(2)`
`1xx785xx10^(-3)=0.052xxV_(2)`
`V_(2)=(1xx785xx10^(-3))/(0.052)`
`=15096.15xx10^(-3) dm^(3)`