We know that pressure inside the curved surface will be greater than of outside pressure of atmosphere.
Let the pressure inside the balloon be $p_i$ and the outside pressure be $\left(p_0\right)$ then excess pressure is
$$
p_i-p_0=\frac{2 \sigma}{r} \text {. }
$$
where, $\sigma=$ Surface tension in membrane of balloon, $r=$ radius of balloon.
Consider, an ideal gas or air inside is perfect then $p_i V=n_i R T_i$
where, $V$ is the volume of the air inside the balloon
$n_i$ is the number of moles inside
$T_i$ is the temperature inside,
So, $n_i=\frac{p_i V}{R T_i}=\frac{\left.M_i \text { (mass of air balloon }\right)}{M_A \text { (molecular mass) }}$
where, $M_i$ is the mass of air inside and $M_A$ is the molar mass of air.
Similarly, $p_0 V=n_0 R T_0$
where $V$ is the volume of the air displaced and $n_0$ is the number of moles displaced and $T_0$ is the temperature outside.
$$
n_0=\frac{p_0 V}{R T_0}=\frac{M_0(\text { mass of displaced air outside })}{M_A(\text { molecular mass })}
$$
where, $n_0=$ no of molecules of air displaced by balloon. $M_0$ is the mass of air outside that has been displaced. By principle of (flotation), if $w$ is the load it can raise, then $w+M_{\mathrm{i}} \mathrm{g}=M_0 g$
Weight lifted by balloon $w=M_0 g-M_i g=\left(M_0-M_i\right) g$
As in atmosphere $21 \% \mathrm{O}_2$ and $79 \% \mathrm{~N}_2$ is present.
$\therefore$ Molar mass of air
$M_i=0.21 \times 32+0.79 \times 28=28.84 \mathrm{~g}$
$\therefore$ Weight raised by the balloon
$$
w=\left(M_0-M_i\right) g
$$
from (i), $M_\phi=\frac{P_i V M_A}{R T_i}$
from (ii), $M_o=\frac{P_0 V M_A}{R T_0}$
$\Rightarrow w=\frac{M_A V}{R}\left(\frac{p_0}{T_0}-\frac{p_i}{T_i}\right) g$
$=\frac{0.02884 \times \frac{4}{3} \pi \times 8^3 \times 9.8}{8.314}$
$\left(\frac{1.013 \times 10^5}{293}-\frac{1.013 \times 10^5}{333}-\frac{2 \times 5}{8 \times 313}\right)$
$=\frac{0.02884 \frac{4}{3} \pi \times 8^3}{8.314} \times 1.013 \times 10^5\left(\frac{1}{293}-\frac{1}{333}\right) \times 9.8$
$=3044.2 \mathrm{~N}$
$\therefore$ Mass lifted by the ballon
$=\frac{w}{g}=\frac{3044.2}{10}=304.42 \mathrm{~kg}$
$=\approx 305 \mathrm{~kg}$.