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Chapter 13: Problem 24

What must be the volume \(V\) of a \(5.0\) -kg balloon filled with helium \(\left(\rho_{\mathrm{He}}=0.178 \mathrm{~kg} / \mathrm{m}^{3}\right)\) if it is to lift a \(30-\mathrm{kg}\) load? Use \(\rho_{\text {air }}=1.29 \mathrm{~kg} / \mathrm{m}^{3}\). The buoyant force, \(V \rho_{\mathrm{air}} g\), must lift the weight of the balloon, its load, and the helium within it: $$ V \rho_{\text {air }} g=(35 \mathrm{~kg})(g)+V \rho_{\mathrm{He}} g $$ which gives $$ V=\frac{35 \mathrm{~kg}}{\rho_{\mathrm{air}}-\rho_{\mathrm{He}}}=\frac{35 \mathrm{~kg}}{1.11 \mathrm{~kg} / \mathrm{m}^{3}}=32 \mathrm{~m}^{3} $$

Short Answer

Expert verified
The volume of the balloon must be approximately 32 \( \mathrm{~m}^3 \).

Step by step solution

01

Understand the Problem

We need to find the volume \( V \) of a helium balloon that can lift a 30-kg load. This includes the weight of the helium itself and the balloon.
02

Identify Given Values

The mass of the balloon is 5.0 kg, the density of helium \( \rho_{\mathrm{He}} = 0.178 \mathrm{~kg/m^3} \), the air density \( \rho_{\text{air}} = 1.29 \mathrm{~kg/m^3} \), and the load to lift is 30 kg.
03

Set Up the Buoyant Force Equation

The buoyant force is given by \( V \rho_{\text{air}} g \). This force must equal the total weight (balloon + helium + load): \( (35 \mathrm{~kg})g + V \rho_{\mathrm{He}} g \).
04

Solve for Volume, \( V \)

Rearrange the equation: \( V \rho_{\text{air}} = 35 \mathrm{~kg} + V \rho_{\mathrm{He}} \). Simplify it to \( V (\rho_{\text{air}} - \rho_{\mathrm{He}}) = 35 \mathrm{~kg} \). Thus, \( V = \frac{35}{\rho_{\text{air}} - \rho_{\mathrm{He}}} \).
05

Substitute Known Values

Substitute \( \rho_{\text{air}} = 1.29 \mathrm{~kg/m^3} \) and \( \rho_{\mathrm{He}} = 0.178 \mathrm{~kg/m^3} \): \( V = \frac{35}{1.29 - 0.178} \).
06

Calculate the Volume

Perform the subtraction: \( 1.29 - 0.178 = 1.112 \mathrm{~kg/m^3} \). Then calculate \( V = \frac{35}{1.112} \approx 31.47 \mathrm{~m^3} \).
07

Check the Calculations

It's important to ensure calculations are consistent with the problem's context; if needed, adjust for significant figures or round as required. Hence, the volume is approximately 32 \( \mathrm{~m^3} \).

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Key Concepts

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

Helium Balloon
A helium balloon is a fascinating object that floats in the air because it is filled with helium gas, which is less dense than the surrounding air. Helium is a noble gas known for its extremely low density and its non-reactive nature. This makes it ideal for creating buoyant balloons. When a balloon is filled with helium, this lighter-than-air gas causes the balloon to rise, opposing the force of gravity.

Helium balloons are commonly used for various applications, such as weather balloons for meteorological observations, or simply for decoration at events. Understanding the science behind why helium balloons float is crucial. It involves examining the gas's properties and comparing it to the air we breathe. By doing so, we reveal the underlying principles of physics that dictate buoyancy and lift.
Density of Gases
Density plays a crucial role in determining whether an object will sink or float. The density of a gas is defined as its mass per unit volume \( ho = \frac{m}{V}\). In the context of helium balloons, we compare the densities of helium and air.

Air has a density of approximately \(1.29\, \text{kg/m}^3\), whereas helium's density is significantly lower, at \(0.178\, \text{kg/m}^3\). This difference is essential for the balloon's ability to lift. Since helium is less dense than air, the balloon experiences an upward force when placed in air, allowing it to float.

Another interesting aspect of gas density is how it changes with temperature and pressure. As temperature increases, or as pressure decreases, the density of a gas reduces slightly, which may influence how much lift a balloon can generate. Comprehending this relationship helps in better understanding how gases behave under different environmental conditions.
Lift Force
Lift force is a critical concept when discussing buoyancy and is what allows objects like helium balloons to float. The lift force is essentially the upward force that counters the weight of the balloon, its payload, and the gas inside. This force can be calculated using the equation for buoyant force, \( ext{Buoyant Force} = V \cdot \rho_{\text{air}} \cdot g\), where \(V\) is the volume, \(\rho_{\text{air}}\) is the air density, and \(g\) is the acceleration due to gravity.

In the case of a helium balloon, the lift force needs to be sufficient to balance not only the mass of the balloon itself but also the added weight of the helium and any additional load suspended from it. If the lift force is greater than the total weight, the balloon will ascend. Otherwise, if it is less, the balloon will not lift off the ground. This simple principle explains the magic of floating balloons!
  • Lift force = Upward force
  • Buoyant advantages = Less dense gases
Archimedes' Principle
Archimedes' Principle is a foundational concept in fluid mechanics, providing the scientific basis for understanding buoyancy. It states that any object fully or partially submerged in a fluid (such as a gas or liquid) experiences a buoyant force equal to the weight of the fluid displaced by the object.

For a helium balloon, this means the balloon experiences enough upward force to counter its weight if the air displaced by the helium balloon weighs more than the combined weight of the balloon and its contents, including the helium.

Archimedes' Principle helps us set the conditions under which a balloon will float. By knowing the densities of both the helium \(\rho_{\text{He}}\) and the air \(\rho_{\text{air}}\), we can use the displacement method to calculate whether the balloon will rise, remain stable, or fall.

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Most popular questions from this chapter

A glass of water has a \(10-\mathrm{cm}^{3}\) ice cube floating in it. The glass is filled to the brim with cold water. By the time the ice cube has completely melted, how much water will have flowed out of the glass? The sp gr of ice is \(0.92\).

A cube of wood floating in water supports a 200-g mass resting on the center of its top face. When the mass is removed, the cube rises \(2.00 \mathrm{~cm}\). Determine the volume of the cube.

A \(60-\mathrm{kg}\) rectangular box, open at the top, has base dimensions of \(1.0 \mathrm{~m}\) by \(0.80 \mathrm{~m}\) and a depth of \(0.50 \mathrm{~m} .(a)\) How deep will it sink in fresh water? \((b)\) What weight \(F_{W b}\) of ballast will cause it to sink to a depth of \(30 \mathrm{~cm}\) ? (a) Assuming that the box floats, $$\begin{array}{c} F_{B}=\text { Weight of displaced water }=\text { Weight of box } \\ \left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m} \times 0.80 \mathrm{~m} \times y)=(60 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) \end{array}$$ where \(y\) is the depth the box sinks. Solving yields \(y=0.075 \mathrm{~m}\). Because this is smaller than \(0.50 \mathrm{~m}\), our assumption is shown to be correct. (b) \(F_{B}=\) weight of box \(+\) weight of ballast But the \(F_{B}\) is equal to the weight of the displaced water. Therefore, the above equation becomes $$\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m} \times 0.80 \mathrm{~m} \times 0.30 \mathrm{~m})=(60)(9.81) \mathrm{N}+F_{W b}$$ from which \(F_{W b}=1760 \mathrm{~N}=1.8 \mathrm{kN}\). The ballast must have a mass of \((1760 / 9.81) \mathrm{kg}=180 \mathrm{~kg}\).

A glass stopper has a mass of \(2.50 \mathrm{~g}\) when measured in air, \(1.50 \mathrm{~g}\) in water, and \(0.70 \mathrm{~g}\) in sulfuric acid. What is the density of the acid? What is its specific gravity? The \(F_{B}\) on the stopper in water is \((0.00250-0.00150)(9.81) \mathrm{N}\). This is the weight of the displaced water. Since \(\rho=m / V\), or \(\rho g=F_{W} / V\), $$\begin{aligned} \text { Volume of stopper } &=\text { Volume of displaced water }=\frac{\text { weight }}{\rho \mathrm{g}} \\ V &=\frac{(0.00100)(9.81) \mathrm{N}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=1.00 \times 10^{-6} \mathrm{~m}^{3} \end{aligned}$$ The buoyant force in acid is $$\left[(2.50-0.70) \times 10^{-3}\right](9.81) \mathrm{N}=(0.00180)(9.81) \mathrm{N}$$ But this is equal to the weight of displaced acid, \(m g\). Since \(\rho=m / V\), and since \(m=0.00180 \mathrm{~kg}\) and \(V=1.00 \times 10^{-6} \mathrm{~m}^{3}\) $$\rho \text { of acid }=\frac{0.00180 \mathrm{~kg}}{1.00 \times 10^{-6} \mathrm{~m}^{3}}=1.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$$ Then, for the acid, $$\text { sp } \mathrm{gr}=\frac{\rho \text { of acid }}{\rho \text { of water }}=\frac{1800}{1000}=1.8$$ Alternative Method $$\begin{array}{l} \text { Weight of displaced water }=\left[(2.50-1.50) \times 10^{-3}\right](9.81) \mathrm{N} \\ \text { Weight of displaced acid }=\left[(2.50-0.70) \times 10^{-3}\right](9.81) \mathrm{N} \end{array}$$ so sp gr of acid \(=\frac{\text { Weight of displaced acid }}{\text { Weight of equal volume of displaced water }}=\frac{1.80}{1.00}=1.8\) Then, since sp gr of acid \(=(\rho\) of acid \() /(\rho\) of water \()\), $$ \rho \text { of acid }=(\text { sp gr of acid })(\rho \text { of water })=(1.8)\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)=1.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} $$

A reservoir dam holds an \(8.00-\mathrm{km}^{2}\) lake behind it. Just behind the dam, the lake is \(12.0 \mathrm{~m}\) deep. What is the water pressure \((a)\) at the base of the dam and \((b)\) at a point \(3.0 \mathrm{~m}\) down from the lake's surface? The area of the lake behind the dam has no effect on the pressure against the dam. At any point, \(P=\rho_{w} g h\). (a) \(P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(12.0 \mathrm{~m})=118 \mathrm{kPa}\) (b) \(P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(3.0 \mathrm{~m})=29 \mathrm{kPa}\)
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