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A baloon filled with helium rises agains...

A baloon filled with helium rises against gravity increasing its potential energy. The speed of the baloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy ? You can neglect viscous drag of air and assume that density of air is constant.

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LET m=Mass of ballon
V=Volume of balloon
`rho_(He)`=Density of helium
`rho_("air")`=Density of air
Volume V of ballon displaces volume V of air.
So, `V(rho_("air")-rho_(He)g=ma=m(dv)/(dt)="up thrust"`
Integrating with respect to t.
`V(rho_("air")-rho_(He)gt=mv`
`(1)/(2)mv^(2)=(1)/(2)m(V^(2))/(m^(2))(rho_("air")-rho_(He))^(2)g^(2)t^(2)`
`(1)/(2m)V^(2)(rho_("air")-rho_(He))g^(2)t^(2)`
If the ballon rises to a height h, from s=`ut+(1)/(2)at^(2)`
`"We get" h=(1)/(2)at^(2)=(1)/(2)(V(rho_("air")-rho_("He")))/(m)"gt"^(2)`
=`V(rho_("air")-rho_("He")gh`
Rearranging the terms
`(1)/(2)mv^(2)+Vrho_("He") gh=V_("pair")hg`
`KE_("ballon")+PE_("ballon")="Change in PE of air"`
So, as the ballon goes up , an equal volume of air comes down, increase in PE and KE of the ballon is the ballon is at the cost of PE of air [Which comes down]
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