Question

Calculate the volume of the balloon filled with hydrogen gas, which will be sufficient to lift a load of $25Kg$ in air. Given that densities of air and hydrogen are $12.9 \times {10^{ - 4}}gc{m^{ - 3}}$ and $9 \times {10^{ - 5}}gc{m^{ - 3}}$ .
(A) $25{m^3}$
(B) $24{m^3}$
(C) $0.283{m^3}$
(D) None of these.

Answer 1

The gas that has no colour, odour and has the densities that is very low than other gases is called Hydrogen. In the periodic table, we will first come across hydrogen. The hydrogen is represented in the symbol $H$ . To solve the given problem, consider the thrust is always greater than the sum of the weights.
$\Rightarrow v{\rho _{air}}g > v\rho g + mg$
Where,
$v$ is the volume, ${\rho _{air}}$ is the density of the air, $g$ is the gravity and $m$ is the mass.

Complete answer:
Consider the given values. It is given to a balloon that is filled with hydrogen. The balloon is sufficient enough to hold $25Kg$ in air. The density values of hydrogen and the air is given. That is,
$12.9 \times {10^{ - 4}}gc{m^{ - 3}}$ and $9 \times {10^{ - 5}}gc{m^{ - 3}}$ respectively.
The volume of the balloon that is filled with the hydrogen is what we need to find.
To solve the given question, remember the condition. The condition is that the given thrust is always greater than the sum of the weights of the hydrogen gas and the balloon. That is,
$\Rightarrow th > {W_H} + {W_B}$
Where,
$th$ is the thrust, ${W_H}$ hydrogen weight and ${W_B}$ balloon weight.
Consider the formula given,
$\Rightarrow v{\rho _{air}}g > v\rho g + mg$
Where,
$v$ is the volume, ${\rho _{air}}$ is the density of the air, $g$ is the gravity and $m$ is the mass.
Take the common terms to the right side.
$\Rightarrow v({\rho _{air}} - {\rho _H})g > mg$
The value of the volume only needs to be found. That is,
$\Rightarrow v = \dfrac{{mg}}{{({\rho _{air}} - {\rho _H})g}}$
Cancel out the common term,
$\Rightarrow v = \dfrac{m}{{({\rho _{air}} - {\rho _H})}}$
We have $m = 25Kg$ convert the mass value into grams. $m = 25 \times {10^3}g$ , ${\rho _{air}} = 12.9 \times {10^{ - 4}}gc{m^{ - 3}}$ and ${\rho _H} = 9 \times {10^{ - 5}}gc{m^{ - 3}}$ . the densities values can be written as, $0.00129$ and $0.0009$ . Substitute all the values.
$\Rightarrow v = \dfrac{{25 \times {{10}^3}}}{{(0.00129 - 0.0009)}}$
Simplify the equation,
$\Rightarrow v = \dfrac{{25 \times {{10}^3}}}{{12}}$
Divide to get the answer.
$\Rightarrow v = 0.2083{m^3}$
Therefore, the volume of the balloon is $0.283{m^3}$ .
Hence $(C)$ is the correct option.

Note:
Hydrogen is the most and first basic element of nature and hence it has various uses. The most significant use of the hydrogen is the ammonia synthesis. In the catalytic hydration of the vegetable oils large amounts of hydrogen is used. It is also famous for its rocket fuels. When combined with the oxygen it is used as the rocket fuel. With the help of nuclear energy it is also used as rocket propellant.