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Mechanical Properties of Fluids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMechanical Properties of Fluids5 Marks
Question
A hot air balloon is a sphere of radius 8 m . The air inside is at a temperature of $60^{\circ} \mathrm{C}$. How large a mass can the balloon lift when the outside temperature is $20^{\circ} \mathrm{C}$ ? (Assume air is an ideal gas, $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}, 1 \mathrm{~atm} .=1.013 \times$ $10^5$ Pa the membrane tension is $5 \mathrm{~N} \mathrm{~m}^{-1}$)
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Answer

Pressure inside $(P_i)$ balloon is larger than outer pressure $(p_0)$ of atmosphere.$\therefore\ \text{P}_\text{i}-\text{P}_0=\frac{2\sigma}{\text{R}}$
$\sigma=$ surface tension in membrane of balloon R = radius of balloon. (image) Gas or air inside is perfect (considered)$\therefore\ \text{P}_\text{i}\text{V}=\text{n}_\text{i}\text{RT}_\text{i}$
V = volume of balloon $n_i$ = no. of moles of gas in balloon R = gas constant $T_i$ = temperature of balloon$\text{n}_\text{i}=\frac{\text{P}_\text{i}\text{V}}{\text{RT}_\text{i}}=\frac{\text{mass of air balloon(M}_\text{i})}{\text{molecular mass (m}_\text{A})}$
$\text{n}_\text{i}=\frac{\text{M}_\text{i}}{\text{M}_\text{A}}=\frac{\text{P}_\text{i}\text{V}}{\text{RT}_\text{i}}$
Similarly, $\text{n}_0=\frac{\text{P}_0\text{V}}{\text{RT}_0}$ By principle off floataion $\text{W}+\text{M}_\text{i}\text{g}=\text{M}_0\text{g}$ W = weight lifted by balloon $\text{W}=\text{M}_0\text{g}-\text{M}_\text{i}\text{g}$$\text{W}=(\text{M}_0-\text{M}_\text{i})\text{g}$
where $n_0$ = no. of molecules of air displaces by balloon. V = Volume of air displced by ballon equal to volume of balloon. If $M_0$ mass of air displaced by balloon. $M_A$ = molecular mass inside or outside balloon.$\therefore\ \text{n}_0=\frac{\text{M}_0}{\text{M}_\text{A}}\text{ or }=\frac{\text{M}_0}{\text{M}_\text{A}}=\frac{\text{P}_0\text{V}}{\text{RT}_\text{0}}$
$\text{M}_0=\frac{\text{P}_0\text{VM}_\text{A}}{\text{RT}_0}$
From (i) $\text{M}_\text{i}=\frac{\text{P}_\text{i}\text{V}\cdot\text{M}_\text{A}}{\text{RT}_\text{i}}$$\therefore\ \text{W}=\Big[\frac{\text{P}_0\text{VM}_\text{A}}{\text{RT}_0}-\frac{\text{P}_\text{i}\text{VM}_\text{A}}{\text{RT}_\text{i}}\Big]\text{g}$
$\text{W}=\frac{\text{VM}_\text{A}}{\text{R}}\Big[\frac{\text{P}_0}{\text{T}_0}-\frac{\text{P}_\text{i}}{\text{T}_\text{i}}\Big]\text{g}$
$\text{M}_\text{A}=21\%\text{O}_2+79\%\text{ or N}_2$
$\text{M}_\text{A}=0.21\times32+0.79\times28=4\big[0.21\times8+0.79\times7\big]$
$\text{M}_\text{A}=4\big[1.68+5.53\big]=4\big[7.21\big]=28.84\text{g}$
$\text{M}_\text{A}=0.2884\text{kg P}_\text{i}=\text{P}_0+\frac{2\sigma}{\text{R}}$
$\text{W}=\frac{\frac{4}{3}\pi\times8\times8\times8\times0.2884}{8.314}\Big[\frac{1.013\times10^5}{(273+20)}-\frac{\text{P}_\text{i}}{273+60}\Big]\text{g}$
$\text{P}_\text{i}=\text{P}_0+\text{P}$ due to S.T of membrane $=\text{p}_0+\frac{2\sigma}{\text{R}}$
$\text{P}_\text{i}=\Big[1.013\times10^5+\frac{2\times5}{8}\Big]=\big[1011300+1.25\big]$
$\text{P}_\text{i}=101301.25=1.0130125\times10^5\cong1.013\times10^5$
$\therefore\ \text{W}=\frac{4\times3.14\times8\times8\times8\times0.02884}{3\times8.314}\Big[\frac{1.013\times10^5}{293}-\frac{1.013\times10^5}{333}\Big]\text{g}$
$\text{W}=\frac{4\times3.14\times8\times8\times8\times0.02884\times1.013\times10^5}{3\times8.314}\Big[\frac{1}{293}-\frac{1}{333}\Big]\text{g}$
$\text{W}=\frac{4\times3.14\times8\times8\times8\times0.02884\times1.013\times10^5\times9.8}{3\times8.314}\Big[\frac{333-293}{293\times333}\Big]$
$\text{W}=\frac{3.14\times64\times32\times0.02884\times1.013\times10^5\times9.8\times40}{3\times8.314\times293\times333}=3044.2\text{N}$

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