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) K^(-1), 1 atm. = 1.013 xx 10^5 Pa` , the membrane tension is `5 Nm^(-1))`

Explanation

Let `P_0 P_i` be the pressure outside and inside the balloon. Then excess pressure inside the balloon is `P_i - P_0 = (2S)/(r ) or P_i = P_0 + (2S)/(r ) …. (i)` where s is the surface tension of membrane. For an ideal gas `P_i = V = mu_i R T_i` where V is the volume of air inside the balloon, `mu_i` is the number of moles of air inside the balloon and `T_i` is the temperature inside the balloon. `mu_1 = (P_iV)/(RT_i) = ("mass of air inside the balloon "(M_i))/("molar mass of air "(M_A)) = (M_i)/(M_A) :' = (M_A P_iV)/(RTi)`
Also `P_0 V = mu_0 RT_0` where v is the volume of air displaced and `mu_0` is the number of moles displaced and `T_0` is the temperature outside the balloon. Then `mu_0 = (P_0V)/(RT_0) = (M_0)/(M_A) = ("mass of air outside that has been displaced "(M_0))/("molar mass of air "(M_A))`
`:. M_0 = (M_A P_0V)/(RT_0)`
weight raised by the balloon, `W = (M_0 = M_i)g` Mass to be lifted, ` m= (W)/(g) = M_0 =M_i` Atmospheric air has 21% `O_2 and 79% N_2.` the molar mass of air, `M_A = 0.21 xx 32 + 0.79 xx 28 = 28.82 g = 28.84 xx10^9-3)kg`
`:. m = (M_AV)/(R ) [(P_0)/(T_0) - (P_i)/(T_i)] = (M_A V)/(R ) [ (P_0)/(T_0) -(1)/(T_i) (P_0 + (2S)/(r ))]`
`=(M_AV)/(R ) [(P_0)/(T_0) - (P_0)/(T_i) - (2S)/(T_i xx r) ] = (M_Axx(4)/(3)pi r^3)/(R )[ P_0((1)/(T_0) - (1)/(T_i)) - (2S)/(rT_i)]`
`=((28.84 xx10^(-3))xx4xx(3.14)xx8^3)/(3xx8.314) xx[1.013 xx 10^5 ((1)/(293) - (1)/(333)) - (2xx5)/(8xx333)] = 304.4kg`