A hot air balloon is a sphere of radius $8$ $m$. The air inside is at a temperature of $60^{°}$ $C$. How large a mass can the balloon lift when the outside temperature is $20^{°}$ $C$ ? Assume air is an ideal gas, $R = 8.314\,J\,mol{e^{ - 1}},1\,atm = 1.013 \times {10^5}{P_a},$ the membrane tension is $= 5\,N/m$.

The pressure inside the balloon be $\mathrm{P}_{i}$ and the outside pressure be $\mathrm{P}_{0}$ and $\mathrm{P}_{i}>\mathrm{P}_{0}$.

Excess pressure $\mathrm{P}_{i}-\mathrm{P}_{\mathrm{o}}=\frac{2 \mathrm{~s}}{r}$

where $\mathrm{s}=$ surface tension, $r=$ radius of balloon

Considering the air to be an ideal gas,

$\mathrm{P}_{i} \mathrm{~V}=n_{i} \mathrm{RT}_{i}$   $.....(1)$

(Where $\mathrm{V}=$ is the volume of the air inside the balloon)

$n_{i}=$ the number of moles of inside air

$\mathrm{T}_{i}=$ the temperature of inside air

For air outside the balloon,

$\mathrm{P}_{\mathrm{o}} \mathrm{V}=n_{\mathrm{o}} \mathrm{RT}_{\mathrm{o}}$     $....(2)$

where $V=$ volume of the air displaced

$\mathrm{P}_{\mathrm{o}}=\text { air pressure outside }$

$n_{\mathrm{o}}=\text { number of moles of air displaced }$

$\mathrm{T}_{\mathrm{o}}=\text { outside temperature }$

From equation $(1)$,

$n_{i}=\frac{\mathrm{P}_{i} \mathrm{~V}}{\mathrm{RT}_{i}}=\frac{\mathrm{M}_{i}}{\mathrm{M}_{\mathrm{A}}}$$.....(3)$

(where $\mathrm{M}_{i}$ mass of air inside and $\mathrm{M}_{\mathrm{A}}$ is the molar mass of air.)

From equation $(2)$,

$n_{0}=\frac{\mathrm{P}_{0} \mathrm{~V}}{\mathrm{RT}_{0}}=\frac{\mathrm{M}_{0}}{\mathrm{M}_{\mathrm{A}}}$$.....(4)$

Where $\mathrm{M}_{0}=$ mass of air outside that has been displaced.)