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Karnataka Board PUCPUC Science Class 11

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?

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Question

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 Nm–1.)

Long Answer
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Solution

The pressure inside `P_i` balloon is larger than the outer pressure `P_a` of the atmosphere.

∴ `P_c = P_a = (2σ)/R`

σ = surface tension in the membrane of balloon R = radius of the balloon.

Gas or air inside is perfect (considered)

∴ `PV = n_iRT_i`

`V` = volume of the balloon

`n_c` = no. of moles of gas in the balloon

`R` = gas constant

`T_i` = temperature of balloon

`n_i = (PV)/(RT_i) = "mass of balloon (M)"/("molecular mass" (M_A))`

`n_i = M_1/M_A = (PV)/(RT_i)`

Similarly, `n_a = (P_aV)/(RT_a)`

By principal off floatation `W + M, g = Mσg`

W = weight lifted by balloon `W = M_ag - M_ig`

W = `(M_a - M_i)g`

Where `n_a` = no. of molecules of air displaced by balloon.

V = volume of air displaced by balloon equal to the volume of balloon If `M_a` mass of air displaced by the balloon

`M_A` = molecular mass inside or outside the balloon

∴ `n_σ = M_a/M_A`

`n_o = M_a/M_A = (P_oV)/(RT_a)`

⇒ `M_a = (P_aVM_A)/(RT_o)`

From (i), `M_i = (P_iVM_A)/(RT_i)`

W = `((P_0VM_A)/(RT_a) - (PVM_A)/(RT_i))g`

w = `(VM_A)/R (P_a/T_a - P_c/T_i)g`

`M_A = 21%` of `O_2 + 79%` of `N_2`

`M_A = 0.21 xx 32 + 0.79 xx 28`

`M_A = 4(0.21 xx 8 + 0.79 xx 7)`

`M_A = 4(1.68 + 5.53)`

`M_A = 4(7.21)`

`M_A = 28.84 g`

`M_A = 0.2884  kg`

`P_i = P_σ + (2σ)/R`

W = `4/3 pi xx 8 xx 8 xx 8 xx 0.2884`

= `[(1.013 xx 10^5)/(273 + 20) - P_i/(273 + 60)]g`

`P_i = P_a + P = P_a + (2σ)/R`

`P_i = [1.013 xx 10^5 + (2 xx 5)/8] = 101300 + 1.25`

`P_i = 101301.25 = 1.0130125 xx 10^5 = 1.013 xx 10^5`

∴ W = `(4 xx 3.14 xx 8 xx 8 xx 8 xx 0.02884)/(3 xx 8.314) [(1.013 xx 10^5)/293 - (1.013 xx 10^5)/333]g`

W = `(4 xx 3.14 xx 8 xx 8  xx 8 xx 0.02884 xx 1.013 xx 10^5)/(3 xx 8.314) [1/293 -1/333]g`

W = `(4 xx 3.14 xx 8 xx 8  xx 8 xx 0.02884 xx 1.013 xx 10^5 xx 9.8)/(3 xx 8.314) [1/293 -1/333]`

W = 

`(4 xx 3.14 xx 8 xx 8  xx 8 xx 0.02884 xx 1.013 xx 10^5 xx 9.8 xx 40)/(3 xx 8.314)`

= 3044.2 N

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Chapter 10: Mechanical Properties of Fluids - Exercises [Page 76]

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NCERT Exemplar Physics Exemplar [English] Class 11
Chapter 10 Mechanical Properties of Fluids
Exercises | Q 10.23 | Page 76

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