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 mole–1K^-1, 1 atm. = 1.013×10⁵ Pa; the membrane tension is 5 N m^-1.)

This is a long answer type question as classified in NCERT Exemplar

Let the pressure inside the ballon be P₁ and the outside pressure be P_o, then excess pressure is P_i-P_o =2S/r

Considering the air to be an ideal gas p_iV = n_iRT_i = where, V is the volume of the air inside the balloon, n_i is the number of moles inside and T_i is the temperature inside, and p_oV =n_oRTo where V is the volume of the air displaced and n_o is the number of moles displaced and T_o is the temperature outside.

So n_{i= $\frac{P_{i}V}{R{T}_i}=\frac{M_i}{M_A}$}

Where M_i is the mass of air inside and M_A is the molar mass of air

n_o= $\frac{P_{o}V}{R{T}_o}=\frac{M_o}{M_A}$

if w Is the load it can raise then w+M₁g=M_og

as atmosphere 21% O₂ and 79%N₂ is pres

...more

This is a long answer type question as classified in NCERT Exemplar

Let the pressure inside the ballon be P₁ and the outside pressure be P_o, then excess pressure is P_i-P_o =2S/r

Considering the air to be an ideal gas p_iV = n_iRT_i = where, V is the volume of the air inside the balloon, n_i is the number of moles inside and T_i is the temperature inside, and p_oV =n_oRTo where V is the volume of the air displaced and n_o is the number of moles displaced and T_o is the temperature outside.

So n_{i= $\frac{P_{i}V}{R{T}_i}=\frac{M_i}{M_A}$}

Where M_i is the mass of air inside and M_A is the molar mass of air

n_o= $\frac{P_{o}V}{R{T}_o}=\frac{M_o}{M_A}$

if w Is the load it can raise then w+M₁g=M_og

as atmosphere 21% O₂ and 79%N₂ is present

so molar mass of air

M_i=0.21 $\times 32+0.79\times 28$ = 28.84g

So weight raised by the ballon

w=(M_o-M_i)g

= $\frac{0.02884\times \frac{4}{3}\pi \times 8^3\times 9.8}{8.314}\left(\frac{1.013\times {10}^5}{293}-\frac{1.013\times {10}^5}{333}-\frac{2\times 5}{8\times 313}\right)$

= 3044.2N

Mass lifted by the ballon =w/g =3044.2/10=304.42kg

So approximately 305kg

less

<p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p>Let the pressure inside the ballon be P₁ and the outside pressure be P_o, then excess pressure is P_i-P_o =2S/r</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745474645phpXc3JKX_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745474645phpXc3JKX.jpeg" alt="" width="170" height="132"></picture></div></div><p>Considering the air to be an ideal gas p_iV = n_iRT_i = where, V is the volume of the air inside the balloon, n_i is the number of moles inside and T_i is the temperature inside, and p_oV =n_oRTo where V is the volume of the air displaced and n_o is the number of moles displaced and T_o is the temperature outside.</p><p>So n_{i=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> </span>}</p><p>Where M_i is the mass of air inside and M_A is the molar mass of air</p><p>n_o=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> </span></p><p>if w Is the load it can raise then w+M₁g=M_og</p><p>as atmosphere 21% O₂ and 79%N₂ is present</p><p>so molar mass of air</p><p>M_i=0.21<span contenteditable="false"> <math> <mo>×</mo> <mn>32</mn> <mo>+</mo> <mn>0.79</mn> <mo>×</mo> <mn>28</mn> </math> </span>= 28.84g</p><p>So weight raised by the ballon</p><p>w=(M_o-M_i)g</p><p>=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>0.02884</mn> <mo>×</mo> <mfrac> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> <mi>π</mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>9.8</mn> </mrow> </mrow> <mrow> <mrow> <mn>8.314</mn> </mrow> </mrow> </mfrac> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1.013</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>293</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <mrow> <mn>1.013</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>333</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <mrow> <mn>2</mn> <mo>×</mo> <mn>5</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> <mo>×</mo> <mn>313</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> </span></p><p>= 3044.2N</p><p>Mass lifted by the ballon =w/g =3044.2/10=304.42kg</p><p>So approximately 305kg</p>

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