Answer
440 K
Work Step by Step
The buoyant force is equal to the total weight of the balloon, at the instant just before the balloon lifts off. Also according to Archimedes' principle, the buoyant force is the weight of the displaced outside air. So, we can write,
$$\rho Vg=m_{t}g-(1)$$
We know that,
Mass of the air = number of moles $\times$ Molecular mass
$m_{air}=nM=(\frac{PV}{RT})M-(2)$
(2)=>(1),
$$\rho V=m_{load}+m_{air}=m_{load}+(\frac{PV}{RT})M$$
$T=\frac{PVM}{(\rho V-m_{load})R}$ ; Let's plug known values into this equation.
$T=\frac{(1.01\times10^{5}Pa)(650\space m^{3})(29\times10^{-3}kg/mol)}{[(1.29\space kg/m^{3})(650\space m^{3})-320\space kg](8.31\space J/mol\space K)}=440\space K$
The temperature of the air = 440 K