Question

A balloon of diameter 20 metre weighs 100 kg. If its pay-load is$\,\text{424}\text{.67 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{x}}}\text{g}$, if it is filled with, He at 1.0 atm and${{27}^{\circ }}C$. Density of air is $1.2kg{{m}^{-3}}$ $\left( R=0.082d{{m}^{3}}atm{{K}^{-1}}mo{{l}^{-1}} \right)$.
Then, the value of x is . . . . . . . . .

Answer 1

Calculate the volume occupied by the volume of given dimensions. Substitute the values given in question to determine the mass of He in the balloon. - Determine the mass of air displaced. With this you can determine the total payload value. The payload value will help you determine the value of x mentioned in the question.

Complete step by step answer:
- So in the question it is asked if a balloon of diameter 20m which has a weight of 100 kg and if its payload is $\,\text{424}\text{.67 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{x}}}\text{g}$ and the balloon is filled with Helium gas at the given pressure and temperature, then find the value of x.
We will first make a note of the data given to us in the question.
The value of radius of balloon $=\,10m$
Weight of the balloon $= 100kg$
$\text{Payload}\,\text{=}\,\text{424}\text{.67 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{x}}}\text{g}$ $\text{Payload}\,\text{=}\,\text{424}\text{.67 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{x}}}\text{g}$
Pressure $=\,1atm$
Temperature $= {{27}^{\circ }}C$
Density of air $= 1.2kg{{m}^{-3}}$
We will now calculate the volume of the balloon.
$\text{Volume}\,\text{=}\,\dfrac{\text{4}}{\text{3}}\text{ }\\!\\!\pi\\!\\!\text{ }{{\text{r}}^{\text{3}}}$
$\text{Volume}\,\text{=}\,\dfrac{\text{4}}{\text{3}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{22}}{\text{7}}{{\left( \dfrac{\text{20}}{\text{2}}\text{ }\\!\\!\times\\!\\!\text{ 100} \right)}^{\text{3}}}$
$\text{Volume}\,\,\text{=}\,\text{4190 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{6}}}\text{c}{{\text{m}}^{\text{3}}}$
$\text{Volume}\,\,\text{=}\,\text{4190 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{3}}}\text{liters}$
$\text{Mass}\,\text{of}\,\text{He}\,\text{in the balloon}\,\text{=}\,\dfrac{\text{PVm}}{\text{RT}}$
$\text{Mass}\,\text{of}\,\text{He}\,\text{in the balloon}\,\text{=}\,\dfrac{\text{1 }\\!\\!\times\\!\\!\text{ 4190 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{3}}}\text{ }\\!\\!\times\\!\\!\text{ 4}}{\text{0}\text{.082 }\\!\\!\times\\!\\!\text{ 300}}$
$\text{Mass}\,\text{of}\,\text{He}\,\text{in the balloon}\,\text{=}\,\text{68}\text{.13 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{4}}}\text{g}$
Thus, the total mass$\text{=}\,\,\text{68}\text{.13 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{4}}}\text{g}\,\text{+ 10}\times \text{1}{{\text{0}}^{4}}g=\,78.13\times {{10}^{4}}g$
Mass of air displaced $=1.2\times 4190kg\,=\,5028kg$
- Payload is defined as the difference of mass of air displaced and the total mass of balloon and gas.
$\text{Payload}\,\text{=}\,\text{mass}\,\text{of}\,\text{air}\,\text{displaced}\,\text{-}\,\text{the}\,\text{mass}\,\text{of}\,\text{(balloon}\,\text{+}\,\text{gas)}\,$
$\text{Payload}\,\text{=}\,\text{502}\text{.8 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{4}}}\text{-78}\text{.13 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{4}}}\,\text{=}\,\text{424}\text{.67 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{4}}}\text{g}$
Equating the power of 10 to x, we get the value of x as 4.
From the above calculation we can conclude that the value of x is 4.

Note: It is important to convert all quantities to their standard units before substituting them in the equation or formula. This is done to avoid making errors in calculation.