A helium ballon occupies 8.0 L at `20^(@)C` and 1.0- atm pressure. The ballon rises to an altitude where the air pressure is 0.65 atm and the temperature is `-10^(@)C`. What is its volume when it reaches equilibrium at the new altitude? Note: Neglect tension forces in te material of the balloon.

Answer

11.05L

Explanation

`(P_(1)V_(1))/(T_1) = (P_(2)V_(2))/(T_2)`
`(P_1) =1atm`
`(V_1) = 8L`
`T_(1) = 20^(@) = 293 K`
`(P_2) =0.65 atm`
`(V_2) = ?`
`T_(2) = - 10^(@)C = 263 K`
`P_(2) = 0.65 atm`
`V_(2) =?`
`T_(2) =-10^(@)`C = 263 K`
`(1xx8)/(293) = (0.65V_(2))/(263)`
`v_(2) =(8xx263)/(293xx0.65)`
`V_(2) = 11.05 L`