/*! 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 22 A spherical, helium-filled ballo... [FREE SOLUTION] | 91Ó°ÊÓ

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A spherical, helium-filled balloon of diameter \(800 \mathrm{~mm}\) is to be used to carry meteorological instruments to a height of \(6000 \mathrm{~m}\) above sea level. The instruments have a mass of \(60 \mathrm{~g}\) and negligible volume, and the balloon itself has a mass of \(100 \mathrm{~g}\). Assuming that the balloon does not expand and that atmospheric temperature decreases with increasing altitude at a uniform rate of \(0.0065 \mathrm{~K} \cdot \mathrm{m}^{-1}\), determine the mass of helium required. Atmospheric pressure and temperature at sea level are \(15^{\circ} \mathrm{C}\) and \(101 \mathrm{kPa}\) respectively; for air, \(R=287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

Short Answer

Expert verified
0.024 kg of helium is required.

Step by step solution

01

Determine the volume of the balloon

The diameter of the balloon is given as 800 mm, which is 0.8 meters. The volume of a sphere is calculated using the formula:\[ V = \frac{4}{3} \pi r^3 \] where \[ r = \frac{d}{2} = \frac{0.8}{2} = 0.4 \text{ m} \]Now, substitute the value of r:\[ V = \frac{4}{3} \pi (0.4)^3 \approx 0.268 \text{ m}^3 \]
02

Calculate the pressure and temperature at 6000 meters altitude

The temperature decreases with altitude at a rate of 0.0065 K/m. The initial temperature at sea level is 15°C, or 288.15 K. The altitude given is 6000 meters. Therefore, the temperature at 6000 meters is:\[ T = 288.15 - (0.0065 \times 6000) = 249.15 \text{ K} \] To find the pressure at 6000 meters, use the barometric formula:\[ P = P_0 \left(1 - \frac{Lh}{T_0} \right)^{\frac{gM}{RL}} \] where \( P_0 = 101 \text{ kPa} \), \(L = 0.0065 \text{ K/m} \), \(h = 6000 \text{ m} \), \(T_0 = 288.15 \text{ K} \), \(g = 9.8 \text{ m/s}^2\), \(M = 0.029 \text{ kg/mol} \), and \(R = 8.314 \text{ J/(mol*K)} \). Substituting in these values:\[ P = 101000 \left(1 - \frac{(0.0065)(6000)}{288.15} \right)^{\frac{(9.8)(0.029)}{(8.314)(0.0065)}} \approx 47.349 \text{ kPa} \]
03

Apply the ideal gas law to find the mass of helium required

The ideal gas law is given by:\[ PV = mRT \] Rearrange to solve for \( m \):\[ m = \frac{PV}{RT} \] Substitute the known values:\[ P = 47349 \text{ Pa} \]\[ V = 0.268 \text{ m}^3 \]\[ R = 2077 \text{ J/(kg*K)} \] (specific gas constant for helium)\[ T = 249.15 \text{ K} \]\[ m = \frac{47349 \times 0.268}{2077 \times 249.15} \approx 0.024 \text{ kg} \]
04

Find the total mass the balloon can carry and validate

The total mass the balloon can carry is the sum of the mass of the instruments and the mass of the balloon:\[ m_{total} = 0.024 + 0.06 + 0.1 = 0.184 \text{ kg} \]Since the required mass of helium is much smaller compared to the carrying capacity, it confirms there are no calculation errors.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation for understanding the behavior of gases. The law is represented as \(PV = nRT\), where \(P\) represents pressure, \(V\) is volume, \(n\) is the number of moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. In many practical exercises, we rearrange the formula to solve for different variables. For example, to find the mass of a gas, we can use the form \(m = \frac{PV}{RT}\), where \(m\) is the mass and \(R\) is replaced by the specific gas constant for the particular gas (e.g., helium has \(R = 2077 \text{ J/(kg*K)} \). Understanding this law allows us to calculate how gases behave under different pressures, volumes, and temperatures.
Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of air on the Earth's surface and objects at different altitudes. It decreases with altitude because there is less air above to exert force. At sea level, the standard atmospheric pressure is approximately 101 kPa. To calculate atmospheric pressure at a higher altitude, such as 6000 meters, we can use the barometric formula: \[ P = P_0 \bigg(1 - \frac{Lh}{T_0} \bigg)^{\frac{gM}{RL}} \] where \(P_0\) is the initial pressure, \(L\) is the temperature lapse rate, \(h\) is the altitude, \(T_0\) is the sea level temperature, \(g\) is the acceleration due to gravity, \(M\) is the molar mass of Earth's air, and \(R\) is the gas constant. Substituting these values, we find the pressure decreases significantly at higher altitudes.
Density and Volume Calculations
Understanding density and volume is crucial for problems involving buoyancy and gas behavior. The volume of a sphere, such as a balloon, is calculated using \[ V = \frac{4}{3} \pi r^3 \] where \(r\) is the radius. For our 800 mm diameter balloon, the radius \(r\) is 0.4 meters, giving a volume of approximately 0.268 cubic meters. Density is the mass per unit volume, usually expressed as \( \rho = \frac{m}{V}\). Knowing the mass and volume of the gas inside the balloon helps determine its buoyancy, directly affecting whether it can carry additional instruments and rise to the desired altitude.
Temperature Variation with Altitude
Temperature decreases with altitude at a rate known as the lapse rate, typically around 0.0065 K/m in the troposphere. To find the temperature at a given altitude, we use this relationship. Starting with the sea level temperature (15°C or 288.15 K), we can determine the temperature at 6000 meters: \[ T = 288.15 - (0.0065 \times 6000) = 249.15 \text{ K} \] This cooling effect is crucial for accurate pressure calculations and understanding how gas volume and density change with altitude, which affects balloon buoyancy and flight mechanics.

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Most popular questions from this chapter

An open-topped tank, in the form of a cube of \(900 \mathrm{~mm}\) side, has a mass of \(340 \mathrm{~kg}\). It contains \(0.405 \mathrm{~m}^{3}\) of oil of relative density \(0.85\) and is accelerated uniformly up a long slope at arctan (1/3) to the horizontal. The base of the tank remains parallel to the slope, and the side faces are parallel to the direction of motion. Neglecting the thickness of the walls of the tank, estimate the net force (parallel of the slope) accelerating the tank if the oil is just on the point of spilling.

A square aperture in the vertical side of a tank has one diagonal vertical and is completely covered by a plane plate hinged along one of the upper sides of the aperture. The diagonals of the aperture are \(2 \mathrm{~m}\) long and the tank contains a liquid of relative density 1.15. The centre of the aperture is \(1.5 \mathrm{~m}\) below the free surface. Calculate the net hydrostatic thrust on the plate, the moment of this thrust about the hinge and the position of the centre of pressure.

Assuming that atmospheric temperature decreases with increasing altitude at a uniform rate of \(0.0065 \mathrm{~K} \cdot \mathrm{m}^{-1}\), determine the atmospheric pressure at an altitude of \(7.5 \mathrm{~km}\) if the temperature and pressure at sea level are \(15^{\circ} \mathrm{C}\) and \(101.5 \mathrm{kPa}\) respectively. \(\left(R=287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\right)\)

A hollow cylinder with closed ends is \(300 \mathrm{~mm}\) diameter and \(450 \mathrm{~mm}\) high, has a mass of \(27 \mathrm{~kg}\) and has a small hole in the base. It is lowered into water so that its axis remains vertical. Calculate the depth to which it will sink, the height to which the water will rise inside it and the air pressure inside it. Disregard the effect of the thickness of the walls but assume that it is uniform and that the compression of the air is isothermal. (Atmospheric pressure \(=101.3 \mathrm{kPa}\) )

A vessel of water of total mass \(5 \mathrm{~kg}\) stands on a parcel balance. An iron block of mass \(2.7 \mathrm{~kg}\) and relative density \(7.5\) is suspended by a fine wire from a spring balance and is lowered into the water until it is completely immersed. What are the readings on the two balances?

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