Question

A balloon filled with helium gas having pressure 2 atm was kept at room temperature (of `27^(@)`C). What will be the resulting temperature if it bursts? (Assume that change occurs reversibly)
`{:((1),-60^(@)C,(2),-45.65^(@)C),((3),-30.15^(@)C,(4),-10.5^(@)C):}`

Answer 1

Birsting of the balloon is an adiabatic process for which
`(P_(1)^(gamma-1))/(T_(1)^(gamma))=(P_(2)^(gamma-1))/(T_(2)^(gamma))or ((P_(1))/(P_(2)))^(gamma-1)=((T_(1))/(T_(2)))^(gamma)`
Helium is a monoatomic gas for which `gamma` = 5/3. After bursting, the pressure of gas enclosed will be lowered to 1 atm.
`((2)/(1))^((5)/(3)-1)=((300)/(T))^((5)/(3))`
Taking log on both sides, we get,
`(5)/(6)"log"(300)/(T)=(2)/(3)log 2`
On solving, we get,
T = 227.35 K
= -`45.65^(@)C`