Question

Payload is defined as the difference between the mass of displace air and the mass of the balloon. Calculate the payload, when a balloon of radium $10m$ of mass $100kg$ is filled with helium at $1.66$ bar at ${27^\circ }C.$
(Density of air$ = 1.2\,kg{m^{ - 3}}$ and $R = 0.083$$\,bar$$d{m^3}{K^{ - 1}}mo{l^{ - 1}}$)

Answer 1

Hint: Calculate the volume of the balloon. Then use the formula of density to calculate the mass of the displaced air. Then use the ideal gas equation to calculate the number of moles of helium gas and use it to calculate the mass of helium gas. Then use this information to calculate the mass of the payload.

Complete step by step solution:
The volume of the balloon is $V = \dfrac{4}{3}\pi {r^3}$
Radius of balloon is $r = 10m$
$d = 1.2kg\,{m^{ - 3}}$
where $d$ is density of air
We know the value $\pi = 3.14$
Firstly, we will find the volume of the balloon
$V = \dfrac{4}{3}\pi {r^3}$
Put a value given in the question.
$ \Rightarrow V = \dfrac{4}{3} \times 3.1416 \times {(10)^3}$
Simplifying it, we get
$ \Rightarrow V = 4186.7{m^3}$
Now, we know that density is the ratio of mass per unit volume. So, the mass of displaced air is obtained from the product of volume and density.
$ \Rightarrow m = V \times d$
where $m$ is mass of the displaced air.
Putting the values of volume of balloon and the density of air, we get
$m = 4186.7 \times 1.2$
$ \Rightarrow m = 5024\,kg$
It is a value of mass of the displaced air.
Now, we will use the ideal gas equation for further calculation.
Ideal gas equation is
$PV = nRT$
Where,
$P$ is pressure
$V$ is volume
$R$ is gas constant
$T$ is temperature
$n$ is the number of mole of gas
By, rearranging the gas equation. We can write, the number of moles of gas present as
$n = \dfrac{{PV}}{{RT}}$
It is given in the question that,
Pressure of the gas is $P = 1.66\,bar$
Gas constant, $R = 0.083$ $bar$$d{m^3}{K^{ - 1}}mo{l^{ - 1}}$
Volume of the gas in the balloon is
$V = 4186.7\,{m^3} = 4186.7 \times {10^3}d{m^3}$
And the temperature is $T = 27{}^{\circ}C = 300K$
By substituting these values and the value of volume calculated in the above equation, we get
$n = \dfrac{{1.66 \times 4186.7}}{{0.083 \times 300}} \times {10^3}$
$n = 279.11 \times {10^3}\,mole$
Now, the molar mass of helium gas is $4\,gmol{e^{ - 1}}$
Therefore, mass of helium gas is
${m_H} = n \times M$
Where,
${m_H}$ is the mass of helium
$M$ is the molar mass of helium
 $ \Rightarrow {m_H} = 279.11 \times 4 \times {10^3}\,g$
$ \Rightarrow {m_H} = 1116.44 \times {10^3}\,g$
$ \Rightarrow {m_H} = 1116.44\,kg$
The mass of a filled balloon is the sum of the mass of the empty balloon and the mass of helium. i.e.
${m_b} = {m_e} + {m_H}$
Where,
${m_b}$ is the mass of filled balloon
${m_e} = 100\,kg$ is the mass of empty balloon
$ \Rightarrow {m_b} = 100 + 1116.44$
$ \Rightarrow {m_b} = 1216.4\,kg$
It is given that the mass of payload is the difference between the mass of displace air and the mass of the balloon. i.e.
\[{m_p} = m - {m_b}\]
Where,
\[{m_p}\] is the mass of payload.
Now, by substituting the values of mass of displaced air and the mass of filled balloon in the above equation, we get
${m_p} = 5024 - 1216.4$
$ \Rightarrow {m_p} = 3807.6\,kg$

Therefore, the mass of the payload is $3807.6\,kg$.

Note:
This question could be confusing due to various parts of the solution. In such a case, do not try to focus on the entire solution simultaneously. First think that we need to calculate the mass of the payload. For that, we need the mass of the filled balloon and the mass of the displaced are. Once, this idea is clear, then solve the three parts separately.
Since, R was in $d{m^3}$, we converted mass in $d{m^3}$ as well.
$1{m^3} = 1000\,d{m^3}$