/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 (a) A spherical balloon with a d... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) A spherical balloon with a diameter of \(8 \mathrm{~m}\) is filled with helium at \(22^{\circ} \mathrm{C}\) and 210 \(\mathrm{kPa}\). Determine the number of moles and the mass of helium in the balloon. (b) When the air temperature in an automobile tire is \(27^{\circ} \mathrm{C},\) the pressure gauge reads \(215 \mathrm{kPa}\). If the volume of the tire is \(0.030 \mathrm{~m}^{3},\) determine the pressure rise in the tire when the air temperature in the tire rises to \(53^{\circ} \mathrm{C}\). (Note: Volume of a sphere is \(\pi D^{3} / 6 ;\) relative molecular mass of He is 4.003 ; universal gas constant, \(R_{\mathrm{u}}\), is 8312 \(\mathrm{J} / \mathrm{kmol} \cdot \mathrm{K} .\) )

Short Answer

Expert verified
(a) Number of moles: 2290.6, mass: 9164 kg; (b) Pressure rise: 27.6 kPa.

Step by step solution

01

Calculate the volume of the balloon

The volume of a sphere is given by the formula \( V = \frac{\pi D^3}{6} \). Substituting \( D = 8 \mathrm{~m} \), we get: \[ V = \frac{\pi (8)^3}{6} = \frac{512 \pi}{6} = 85.333 \pi \approx 268.08 \, \mathrm{m}^3 \].
02

Calculate the number of moles of helium

Using the ideal gas law \( PV = nRT \), where \( P = 210 \, \mathrm{kPa} = 210000 \, \mathrm{Pa} \), \( V \approx 268.08 \, \mathrm{m}^3 \), \( R = \frac{8312}{4.003} \, \mathrm{J} / \mathrm{kg} \cdot \mathrm{K} \), and \( T = 22^{\circ} \mathrm{C} = 295.15 \, \mathrm{K} \), we can rearrange to find \( n \): \[ n = \frac{PV}{RT} = \frac{210000 \times 268.08}{8312 \times 295.15} \approx 2290.6 \text{ moles} \].
03

Calculate the mass of helium

The mass \( m \) of helium can be found using \( m = n \times M_r \), where \( M_r = 4.003 \text{ kg/kmol} \). Thus, \[ m = 2290.6 \times 4.003 \approx 9164 \text{ kg} \].
04

Calculate the initial absolute pressure in the tire

Given the gauge pressure \( P_g = 215 \mathrm{kPa} \), the absolute pressure is \( P_1 = P_g + 101.3 \mathrm{kPa} = 316.3 \mathrm{kPa} \).
05

Use the ideal gas law to find the final pressure

For the tire, using \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), solve for \( P_2 \), given \( T_1 = 27^{\circ} \mathrm{C} = 300.15 \, \mathrm{K} \) and \( T_2 = 53^{\circ} \mathrm{C} = 326.15 \, \mathrm{K} \). \[ P_2 = P_1 \frac{T_2}{T_1} = 316.3 \times \frac{326.15}{300.15} \approx 343.9 \mathrm{kPa} \].
06

Calculate the pressure rise due to temperature increase

The pressure rise is simply \( P_2 - P_1 = 343.9 - 316.3 = 27.6 \mathrm{kPa} \).

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Key Concepts

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

Helium gas properties
Helium is a fascinating gas with unique properties that make it useful in many applications. It is a noble gas, which means it does not react easily with other elements. Here are some key properties of helium:
  • It is the second lightest and second most abundant element in the universe, following hydrogen.
  • Helium is colorless, odorless, tasteless, and non-toxic, making it safe for use in various settings.
  • Its boiling and melting points are lower than any other element, which means it remains a gas under most conditions.
  • Helium is often used in balloons and airships due to its low density and non-flammable nature compared to hydrogen.
Because helium doesn’t freeze or react like other gases, it plays a crucial role in scientific experiments and industrial processes as well.
Volume of a sphere
To determine the amount of space occupied by a sphere, we use the formula for the volume of a sphere: \[ V = \frac{\pi D^3}{6} \]where \( V \) is the volume, and \( D \) is the diameter of the sphere. Simple as that!
For example, if you have a balloon with a diameter of 8 meters, just plug in the value to find its volume.
Upon calculation, you’ll find that the volume comes out to approximately 268.08 cubic meters, which indicates how much space the helium gas occupies in the balloon.
This method is essential for tasks involving spherical objects in various fields like physics, engineering, and even everyday applications.
Pressure and temperature relationship
The pressure of a gas is intricately linked to its temperature, according to the ideal gas law, which is a convenient model for gases under many conditions. The ideal gas law is represented as:\[ PV = nRT \]where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.
When analyzing how changes in temperature affect pressure, use the relationship \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). This indicates that if the volume and number of moles remain constant, any change in temperature will directly affect the pressure of the gas.
  • Increasing the temperature will cause the gas molecules to move more energetically, leading to higher pressure.
  • Conversely, a decrease in temperature will result in a pressure drop.
This principle is fundamental in understanding how gases behave under varying conditions, such as in an automobile tire or weather balloons.
Molar mass calculation
Calculating molar mass is critical when dealing with gases and performing conversion between different quantity measures. Here’s how to compute the mass of helium in a balloon:First, determine the number of moles \( n \) present using the ideal gas law: \[ n = \frac{PV}{RT} \]Given in the problem, you have a pressure, volume, temperature, and the specific gas constant value set for helium via the universal gas constant.
Once you know how many moles are present, calculate the mass \( m \) using the molar mass \( M_r \) of helium:\[ m = n \times M_r \]Where \( M_r = 4.003 \text{ kg/kmol} \).
In the provided example, we calculated this to find that the mass of helium is approximately 9164 kg.
Understanding this procedure is valuable in fields such as chemistry and engineering, where accurate mass determination of gases is key for process calculations and safety considerations.

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Most popular questions from this chapter

A cylinder contains \(0.3 \mathrm{~m}^{3}\) of air at \(20^{\circ} \mathrm{C}\) and \(120 \mathrm{kPa}\) pressure. With a piston mechanism, the air in the cylinder is compressed isentropically to a pressure of \(700 \mathrm{kPa}\). What is the temperature of the air after compression?

Before embarking on a 40-minute drive on the highway, a motorist adjusts his tire pressure to \(207 \mathrm{kPa}\). At the end of his trip, the motorist measures his tire pressure as \(241 \mathrm{kPa}\). Assume that the volume of the tire remains constant during the trip. (a) Estimate the percentage increase in the temperature of the air in the tire. (b) If the initial temperature of the air in the tire is assumed to be equal to the ambient air temperature of \(25^{\circ} \mathrm{C},\) what is the estimated temperature of the air in the tire at the end of the trip.

A plastic container is completely filled with gasoline at \(15^{\circ} \mathrm{C}\). The container specifications indicate that it can endure a \(1 \%\) increase in volume before rupturing. Within the temperatures likely to be experienced by the gasoline, the average coefficient of volume expansion of gasoline is \(9.5 \times 10^{-4} \mathrm{~K}^{-1}\). Estimate the maximum temperature rise that can be endured by the gasoline without causing a rupture in the storage container.

Consider the case of an air bubble released from the bottom of a \(12-\mathrm{m}\) -deep lake as shown in Figure 1.21 . The bubble is filled with air, it has an initial diameter of \(6 \mathrm{~mm},\) no air is lost or gained in the bubble as it rises, and the air in the bubble and the surrounding lake water maintain a constant temperature of \(20^{\circ} \mathrm{C}\). Atmospheric pressure, \(p_{\mathrm{atm}},\) on the surface of the lake is \(101.3 \mathrm{kPa},\) and the pressure, \(p,\) at any depth, \(z\), below the surface of the lake is given by \(p=p_{\text {atm }}+\gamma z\), where \(\gamma\) is the specific weight of the water in the lake. Estimate the diameter of the bubble when it surfaces.

If pure oxygen is compressed such that its density and pressure are \(5 \mathrm{~kg} / \mathrm{m}^{3}\) and \(450 \mathrm{kPa}\), respectively, estimate the temperature of the oxygen.

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