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A hot air balloon is a sphere of radius 8m . 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 =8.314J" mole"^(-1)k^(-1),1atm=1.013xx10^(5)Pa, the membrane tension is =5N//m. |
Answer» Solution :The pressure inside the balloon be `P_(i) andP_(i)gtP_(o)`. Excess pressure `P_(i)-P_(o)=(2s)/(R )` where s= surface tension ,r =radius of balloon Considering the air to be ideal gas , `P_(i)V=n_(i)RT_(i)`...(1) Where V= is the volume of the air inside the balloon) `n_(i)` = the NUMBER of moles of inside air `T_(i)=` the temperature of inside air For air outside the balloon, `P_(o)V=n_(o)RT_(o)`....(2) where V= volume of the air displaced `P_(o)=` air pressure outside `n_(o)` = number of moles of air displaced `T_(o)=` outside temperature From equation (1), `n_(i)=(P_(i)V)/(RT_(i))=(M_(i))/(M_(A))`.....(3) where `M_(i)` mass of air inside and `M_(A)` is the molar mass of air .) From equation (2), `n_(o)=(P_(o)V)/(RT_(o))=(M_(o))/(M_(A))`....(4) (Where `M_(o)=` mass of air outside that has been displaced.) If W is the load it can raise , then `W+M_(i)g=M_(o)g` `W=M_(o)g-M_(i)g` ...(5)  There is `21%O_(2)and79%N_(2)`in air . `therefore` Molar mass of air `=(0.21xx32)+(0.79xx28)` `=28.84g` Weight carried by balloon , `W=(M_(o)-M_(i))g` From equation (3),(4) putting value of `M_(o)M_(i)` in above equation, `W=((M_(A)V)/(R)(P_(o))/(T_(o))-(M_(A)V)/(R)(P_(i))/(T_(i)))g` `W=(M_(A)V)/(R)[(P_(o))/(T_(o))-(P_(i))/(T_(i))]-g` (Volume of balloon `V=(4)/(3)pir^(3))` `W=(0.02884xx(4)/(3)pixx8^(3)xx9.8)/(8.314)` `[(1.013xx10^(5))/(293)-(1.013xx10^(5))/(333)]` `W=(0.0884xx(4)/(3)pixx8^(3))/(8.314)xx(1.013xx10^(5))` `[(1)/(293)-(1)/(333)]xx9.8` `W=3044.2N` `thereforemg=3044.2` `thereforem=(W)/(g)=(3044.2)/(10)=304.42kg` `m=305kg` |
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