Problem 135 - A Complete Resource Book in Chemistry for JEE Main

Question

A man weigh \(72.15 \mathrm{~kg}\) and want to fly in the sky with the aid of balloons itself weighing \(20 \mathrm{~kg}\) and each containing 50 moles of \(\mathrm{H}_{2}\) gas at \(0.05 \mathrm{~atm}\) and \(27^{\circ} \mathrm{C}\). If the density of air at the given conditions is \(1.25 \mathrm{~g} / \mathrm{L}\), how many such types of balloons he is needed to fly in the sky.

Step-by-Step Solution

Answer

To fly, the man needs at least 3 balloons.

Step 1: Understand Buoyancy Requirement

To lift off the ground, the total upward force provided by the balloons must be greater than the weight of the man and the balloons themselves. This upward force is provided by the buoyant force due to the displaced air.

Step 2: Compute Weight of Man and Balloons

The total weight is the sum of the man's weight and the balloons' weight: \[\text{Total weight} = 72.15 \text{ kg (man)} + 20 \text{ kg (balloons)} = 92.15 \text{ kg}.\]

Step 3: Convert Weight to Newtons

To calculate force in Newtons, use the acceleration due to gravity, \(g = 9.8 \text{ m/s}^2\):\[\text{Total force} = 92.15 \times 9.8 \approx 902.07 \text{ N}.\]

Step 4: Determine Conditions for Buoyancy

Each balloon must carry a load equal to its own weight minus the load lifted given by buoyant force. The density of air being \(1.25 \text{ g/L}\) is needed to find the force correction.

Step 5: Calculate Buoyant Force per Balloon

For each balloon containing \(50\) moles of \(\text{H}_2\), use the ideal gas law, \(PV = nRT\), to find volume:\[V = \frac{nRT}{P},\] where \(R = 0.0821 \text{ L atm/mol K}\), \(T = 300K\), \(P = 0.05 \text{ atm}\), \(n = 50\).Calculate:\[V = \frac{50 \times 0.0821 \times 300}{0.05} = 24630 \text{ L}.\]Convert to \(\text{m}^3:\)\[V = 24.63 \text{ m}^3.\]Buoyant force = Volume * density * g = \(24.63 \text{ m}^3 \times 1.25 \text{ kg/m}^3 \times 9.8 \approx 302.25 \text{ N} \).

Step 6: Determine Number of Balloons Required

Each balloon provides an upward force of \(302.25 \text{ N}\). To counteract the weight of \(902.07 \text{ N}\), determine the number of balloons required:\[n = \frac{902.07}{302.25} \approx 2.99.\]Since you cannot have a fraction of a balloon, round up to \(3\) balloons.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in thermodynamics, serving as a useful tool for understanding the behavior of gases. The formula is given by \[ PV = nRT \], where:

• \( P \): Pressure of the gas
• \( V \): Volume of the gas
• \( n \): Number of moles
• \( R \): Ideal Gas Constant (approximately \( 0.0821 \text{ L atm/mol K} \))
• \( T \): Temperature in Kelvin

To solve problems involving gases, it is essential to know the conditions of the gas, such as pressure, volume, and temperature. In this exercise, the volume of the gas in the balloon was calculated using this law, assuming the behavior of the gas is 'ideal', meaning it follows the laws without deviations. This calculated volume contributes significantly to determining how much buoyancy the balloon can provide, as seen in our solution.

Density Calculation

Density \( \rho \) is a measure of how much mass is contained in a given volume. It is defined by the formula: \[ \rho = \frac{m}{V} \], where:

• \( \rho \): Density
• \( m \): Mass
• \( V \): Volume

The density of a substance can dramatically affect buoyancy, as less dense objects tend to float in a more dense fluid. In this exercise, the density of the displaced air was used to calculate the buoyant force exerted by the air on the balloon (and hence on the man and the balloons themselves). Given the air density of \( 1.25 \text{ g/L} \), we calculated how this density combined with the balloon volume contributes to the buoyant lift necessary for levitation.

Newton's Laws

Newton's Laws of Motion form the foundation of classical mechanics and are crucial for solving physics problems involving forces:

• First Law (Inertia): An object will remain at rest, or in uniform motion, unless acted upon by a force.
• Second Law (F = ma): The force (\( F \)) acting on an object is equal to the mass (\( m \)) of that object times its acceleration (\( a \)).
• Third Law (Action-Reaction): For every action, there is an equal and opposite reaction.

In our exercise, we applied Newton's Second Law primarily. This was evident when calculating the total downward force (weight) acting on both the man and the balloons. By using the weight (a force) and the acceleration due to gravity (\( 9.8 \text{ m/s}^2 \)), we derived Newtons (\( N \)) as the unit of force, which is crucial for determining how many balloons are needed to produce a greater opposing buoyant force. By ensuring sufficient buoyant force, we achieved the lift necessary for flight, showcasing the interplay of forces in a practical context.