Missing question: "what is the density of the hot air?"
Solution:
The balloon is floating at constant height, and this means that the two forces acting on it (the weight and the buoyant force) are in equilibrium:
[tex]W=B[/tex]
where W is the weight of the balloon, which is sum of the weight of the balloon structure and of the hot air inside the balloon, and B is the buoyant force.
The buoyant force is given by:
[tex]B=Vd_ag[/tex]
where V is the volume of the balloon, d is the air density and g is the gravitational acceleration. Plugging numbers into the equation, we find
[tex]B=Vd_ag=(2310 m^3)(1.22 kg/m^3)(9.81 m/s^2)=2.76 \cdot 10^4 N[/tex]
This is equal to the weight of the balloon+hot air inside it:
[tex]2.76 \cdot 10^4 N = W = W_b + W_h[/tex]
where
[tex]W_b[/tex] is the weight of the balloon
[tex]W_h[/tex] is the weight of the hot air inside the balloon
The weight of the balloon is
[tex]W_b = mg = (326 kg)(9.81 m/s^2)=3199 N[/tex]
Which means that the weight of the hot air is
[tex]W_h = W-W_b = 2.76 \cdot 10^4 N - 3199 N =2.44 \cdot 10^4 N[/tex]
which corresponds to a mass of
[tex]m_h = \frac{W_h}{g}= \frac{2.44 \cdot 10^4 N}{9.81 m/s^2}=2487 kg [/tex]
And since the mass is the product between density and volume, we can find the density of the hot air:
[tex]d_h = \frac{m_h}{V}= \frac{2487 kg}{2310 m^3}=1.08 kg/m^3 [/tex]
