Question

A balloon containing one mole of air at \[{\text{1 atm}}\] pressure initially is filled further with air till the pressure is increased to \[{\text{4 atm}}\]. The initial diameter of the balloon is 1 m and the pressure at each stage is proportional to the diameter of the balloon. How many moles of air need to be added to change the pressure from \[{\text{1 atm}}\]to\[{\text{4 atm}}\].

Answer 1

Hint: This problem is based on the ideal gas law equation which connects the pressure and the volume of the gas with the number of moles of the gas and the temperature of the gas in the Kelvin scale. We shall use the ideal gas equation and substitute the value of pressure and volume and use that relation to give two equations. The ratio of these equations can be used to find the moles of gas.

Formula Used:
${\text{V = }}\dfrac{{\text{1}}}{{\text{6}}}{{\pi }}{{\text{d}}^{\text{3}}}$
where d is diameter of the balloon and V is the volume
\[{\text{P = kd}}\]
Where P is pressure, d is the diameter and k is proportionality constant
${\text{PV = nRT}}$

Complete step by step answer:
As per the given problem, the pressure at each stage is proportional to the diameter of the balloon. If the pressure of the gas is P, the diameter is equal to d, then
\[{\text{P}} \propto {\text{d}}\], or, \[{\text{P = kd}}\], where “k” is the proportionality constant between P and d,
${\text{k = }}\dfrac{{{\text{1 atm}}}}{{{\text{1 metre}}}}$
As per the ideal gas law equation,
${\text{PV = nRT}}$,
Where V is the volume of the gas, n is the number of moles of the gas, R is universal gas constant, and T is the temperature of the gas in absolute scale or the Kelvin scale.
The volume of the balloon can be expressed as ${\text{V = }}\dfrac{{\text{1}}}{{\text{6}}}{{\pi }}{{\text{d}}^{\text{3}}}$
Therefore, replacing the values of the pressure and the volume in the gas equation we get,
$\left( {{\text{kd}}} \right){\text{ }}\dfrac{{\text{1}}}{{\text{6}}}{{\pi }}{{\text{d}}^{\text{3}}}{\text{ = nRT}}$
Let the initial diameter of the balloon be ${{\text{d}}_{\text{1}}}$and the initial number of moles of the gas = ${{\text{n}}_{\text{1}}}$therefore the final diameter of the gas = ${{\text{d}}_2}$and final number of moles of the gas = ${{\text{n}}_2}$
Therefore, the ratio for the two gas equations can be expressed as,
 $\dfrac{{{{\text{d}}_{\text{1}}}^4}}{{{{\text{d}}_2}^4}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{1}}}}}{{{{\text{n}}_{\text{2}}}}}$, ${{\text{n}}_{\text{1}}}$= 1 mole of gas and${{\text{d}}_{\text{1}}}$= 1 metre, while${{\text{d}}_2}$= 4 metre as the diameter of the gas increases proportionately with the pressure of the gas.
Now, $\dfrac{{{{\left( 1 \right)}^4}}}{{{{\left( 4 \right)}^4}}}{\text{ = }}\dfrac{1}{{{{\text{n}}_{\text{2}}}}}$
Therefore, the number of moles of the gas when the pressure of the gas is \[{\text{4 atm}}\]is 256 and the extra number of moles of gases that was added is $256 - 1 = 255$moles.

Note:
The combined gas law equation is applicable only to ideal gases and is based on the laws given by Boyle, Charles, and Avogadro. These laws consider that the volume of the gas is very large in comparison to the volume of the gas molecules and that there exists no force of interaction between the gas molecules.