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Chapter 3: Problem 89

A balloon having weight \(50 \mathrm{~kg}\) is filled with \(685.2 \mathrm{~kg}\) of helium gas at \(760 \mathrm{~mm}\) pressure and \(25^{\circ} \mathrm{C}\). What will be its pay load if it displaces \(5108 \mathrm{~kg}\) of air? (a) \(4372.8 \mathrm{~kg}\) (b) \(4392.6 \mathrm{~kg}\) (c) \(4444.4 \mathrm{~kg}\) (d) \(3482.9 \mathrm{~kg}\)

Short Answer

Expert verified
The payload is 4372.8 kg, so option (a) is correct.

Step by step solution

01

Understanding the Problem

To find the payload of the balloon, we need to calculate the difference between the weight of the air displaced by the balloon and the combined weight of the balloon and the helium gas.
02

Calculate Total Weight of the Balloon including Helium

The total weight of the balloon system is the sum of its own weight and the weight of the helium it is filled with. Hence, we have:\[\text{Total weight} = 50 \text{ kg (balloon)} + 685.2 \text{ kg (helium)} = 735.2 \text{ kg}\]
03

Calculate Pay Load

The payload of the balloon is the difference between the weight of the air it displaces and its own total weight:\[\text{Payload} = \text{Weight of air displaced} - \text{Total weight of balloon including helium}\]\[\text{Payload} = 5108 \text{ kg} - 735.2 \text{ kg} = 4372.8 \text{ kg}\]
04

Choose the Correct Answer

The calculated payload is \(4372.8 \text{ kg}\). Thus, the correct answer is option (a).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Buoyancy
Buoyancy is a fascinating force that allows objects to float or rise in a fluid. This concept is crucial for understanding how balloons work, especially when filled with gases lighter than air, like helium. According to Archimedes' principle, an object submerged in a fluid experiences an upward force equal to the weight of the fluid it displaces.

This means if a balloon displaces a certain weight of air, the buoyant force will be equivalent to this weight. The balloon will float if this buoyant force exceeds the weight of the balloon and its contents. Understanding buoyancy helps us determine how much weight a balloon can lift, which is referred to as the payload.
Weight of Displaced Fluid
The weight of the fluid displaced by a balloon is central to calculating its payload. When a balloon is inflated, it pushes away a volume of air corresponding to the space it occupies. The weight of this displaced air is what contributes to the buoyant force.

By calculating this weight, you can determine how much heavier or lighter the combined weight of the balloon and its contents can be for the balloon to float. In the given exercise, the balloon displaces air weighing 5108 kg. This displacement creates the necessary buoyant force allowing the balloon to carry additional weight, such as equipment or passengers. The difference between the weight of the displaced fluid and the balloon's total weight gives us the potential payload.
Helium Properties
Helium is a light, non-flammable gas often used to fill balloons for lift. Being significantly lighter than air, it adds buoyancy, helping balloons rise. Helium, a noble gas, has low density, which contributes to this lifting effect.

Its non-reactive nature makes it safe for use even in various atmospheric conditions. While it does not contribute much to weight, its volume in the balloon determines how much air can be displaced. Therefore, more helium means more displaced air and a larger buoyant force. Understanding the properties of helium is vital when figuring out how much weight a balloon can lift successfully.
Pressure and Temperature Conditions
Conditions like pressure and temperature can significantly impact a balloon's performance. As pressure increases or decreases, so does the gas volume within the balloon, influencing the amount of air displaced.

Temperature also plays a role; warmer temperatures make helium expand while cooler temperatures might contract it. According to the ideal gas law, the volume of the gas is directly proportional to its temperature when pressure is constant. In the initial problem statement, the balloon's helium is subject to 760 mm pressure and 25掳C temperature. These specific conditions help define its lift capacity because they establish how much helium will fill the balloon and thus how much air will be displaced.
  • Helium expands when heated, increasing buoyancy.
  • Lower temperatures condense helium, potentially decreasing lift.
  • Constant volume changes with atmospheric pressure alterations.
All these factors must be considered to achieve accurate payload calculations for any helium-filled balloon.

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Most popular questions from this chapter

A solution has a 1: 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at \(20^{\circ} \mathrm{C}\) are \(400 \mathrm{~mm} \mathrm{Hg}\) for pentane and \(120 \mathrm{~mm} \mathrm{Hg}\) for hexane. The mole fraction of pentane in the vapour phase would be (a) \(0.200\) (b) \(0.549\) (c) \(0.786\) (d) \(0.478\)

'a' and 'b' are van der Wall's constants for gases. Chlorine is more easily liquefied than ethane because [2011] (a) a for \(\mathrm{Cl}_{2}<\mathrm{a}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\) but \(\mathrm{b}\) for \(\mathrm{Cl}_{2}>\mathrm{b}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\) (b) a and \(\mathrm{b}\) for \(\mathrm{Cl}_{2}<\mathrm{a}\) and \(\mathrm{b}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\) (c) a and \(\mathrm{b}\) for \(\mathrm{Cl}_{2}>\mathrm{a}\) and \(\mathrm{b}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) a for \(\mathrm{Cl}_{2}>\mathrm{a}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\) but \(\mathrm{b}\) for \(\mathrm{Cl}_{2}<\mathrm{b}\) for \(\mathrm{C}_{2} \mathrm{H}_{6}\)

In the given sample of a gas all molecules do not possess same speed. Due to frequent molecular collisions, the molecules move with ever changing speeds and also with changing direction. There are three types of velocities (i) root mean square velocity (ii) average velocity and (iii) most probable velocity. By how may folds the temperature of the gas would increase when the RMS velocity of gas molecules in a container of fixed volume is increased from \(5 \times 10^{4} \mathrm{~cm}\) \(\sec ^{-1}\) to \(10 \times 10^{4} \mathrm{~cm} \sec -1 ?\) (a) 6 times (b) 4 times (c) 2 times (d) 8 times

If two moles of an ideal gas at a temperature \(546 \mathrm{~K}\), occupy a volume of \(44.8\) litres its pressure must be (a) \(4 \mathrm{~atm}\) (b) \(3 \mathrm{~atm}\) (c) \(2 \mathrm{~atm}\) (d) \(1 \mathrm{~atm}\)

Which one of the following statement is not true about the effect of an increase in temperature on the distribution molecular speeds in a gas? (a) the most probable speed increases (b) the fraction of the molecules with the most probable speed increases (c) the distribution becomes broader (d) the area under the distribution curve remains the same as the under the lower temperature
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