Let m , V ρ_{He} denote respectively the mass, volume and density of helium baloon and ρ_{air} be density of air

Volume V of baloon displaces volume V of air.

If the baloon rises to a height h, from s = ut + (1/2)at^{2}, we, geAt h = (1/2)at^{2}

So, as the baloon goes up, an equal volume of air comes down, increase in PE and KE of the baloon is at the cost of PE of air [which comes down].