Lecture 2

 Case 1: [Helium Gas]

We know, $$ w = mg = \frac{m}{v}vg = \rho vg = g \rho v $$ 

Now, Net Force , \begin{aligned}F _{Net} & = F - W \\ & = \rho _{air}Vg -  \rho _{He}Vg\\& = gV\rho _{air} \left (1 - \frac{\rho _{He}}{\rho _{air}}\right)...............(1) \end{aligned}

We also know,

\begin{aligned} PV & = nRT \\ & = \frac{m}{M} RT \\ & = \frac{mRT}{M} \\ \implies P & = \frac{m}{V}.\frac{RT}{M} \\ & = \rho . \frac{RT}{M} \\ \therefore \rho & = \frac{PM}{RT} \end{aligned}

Now, if the temperature is constant. [assumption]

\begin{aligned} \therefore \rho_{He} = \frac{P_He}{RT} ~~~and ~~~ \rho_{air} = \frac{PM_{air}}{RT} \end{aligned}

Putting these in (1),

\begin{aligned} F_{Net} & = Vg\rho_{air} \left(1 - \frac{M_{He}}{M_{air}}\right) \\ & = Vg\rho_{air} \left(1 - \frac{4}{28}\right) \\ [we ~know ~&that, M_{He}= 4 ~and ~M_{air}= 28] \end{aligned}

    Here, if we consider the values, we get positive (+)ve value. Therefore the direction of \(F_{Net}\) will be the same as \(F\) because of the same sign (+)ve. Hence the force, \(F_{Net}\) will be in upward direction. Thus the ballon will fly.