/*! 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 76 The surface of the sun has a tem... [FREE SOLUTION] | 91Ó°ÊÓ

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The surface of the sun has a temperature of about 5800 \(\mathrm{K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is 1.67 \(\times 10^{-27} \mathrm{kg} . )\) (b) The escape speed for a particle to leave the gravitational influence of the sun is given by \((2 G M / R)^{1 / 2}\) , where \(M\) is the sun's mass, \(R\) its radius, and \(G\) the gravitational constant (see Example 12.5 of Section \(12.3 ) .\) Use the data in Appendix \(\mathrm{F}\) to calculate this escape speed. (c) Can appreciable quantities of hydrogen escape from the sun? Can any hydrogen escape? Explain.

Short Answer

Expert verified
RMS speed is much lower than escape speed; hydrogen cannot escape the sun significantly.

Step by step solution

01

Understanding the RMS Speed Formula

The root mean square (rms) speed of particles in a gas can be found using the formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant \( 1.38 \times 10^{-23} \text{ J/K} \), \( T \) is the temperature in Kelvin, and \( m \) is the mass of the particle. In this problem, \( m = 1.67 \times 10^{-27} \text{ kg} \) and \( T = 5800 \text{ K} \).
02

Calculate the RMS Speed of a Hydrogen Atom

Substitute the given values into the rms speed formula:\[ v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 5800}{1.67 \times 10^{-27}}} \]Calculate \( v_{rms} \) to find the speed of hydrogen atoms on the sun's surface.
03

Evaluating the Expression for the Escape Speed

The escape speed \( v_{escape} \) is given by \( v_{escape} = \sqrt{\frac{2GM}{R}} \). Using Appendix F, we find the following constants: \( G = 6.674 \times 10^{-11} \text{ m}^3\text{kg}^{-1}\text{s}^{-2} \), \( M = 1.989 \times 10^{30} \text{ kg} \), and \( R = 6.96 \times 10^{8} \text{ m} \).
04

Calculate the Escape Speed from the Sun

Substitute the known values into the escape speed equation:\[ v_{escape} = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 1.989 \times 10^{30}}{6.96 \times 10^{8}}} \]Calculate \( v_{escape} \) to determine the speed needed to escape the Sun's gravitational pull.
05

Comparing RMS and Escape Speeds

Compare the calculated rms speed to the escape speed. Determine whether the rms speed is sufficient for hydrogen atoms to escape the sun's gravitational influence, and if not, whether some could still escape due to high-speed tail effects.

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Key Concepts

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

RMS Speed
The concept of root mean square (RMS) speed is vital in understanding how molecules move in a gas. RMS speed gives us an average measure of the speed of particles in a gas at a given temperature. It's derived from the kinetic theory of gases, implying that each particle doesn't travel at the same speed but has a distribution of speeds.
To find RMS speed, we use the formula:
  • \[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where:
  • \( k \) is the Boltzmann constant, \( 1.38 \times 10^{-23} \text{ J/K} \)
  • \( T \) is the temperature in Kelvin
  • \( m \) is the mass of a single particle
In this context, for hydrogen atoms on the surface of the sun, we substitute the known values. The mass of a hydrogen atom is \( 1.67 \times 10^{-27} \text{ kg} \) and the temperature \( T \) is \( 5800 \text{ K} \). By plugging these into the formula above, we can calculate the RMS speed specific to a hydrogen atom at these conditions.
Escape Velocity
Understanding escape velocity is crucial when discussing how objects or particles can overcome a celestial body's gravitational pull. Escape velocity is the minimum speed needed for an object to "break free" from the gravitational attraction without further propulsion.
It is calculated using the formula:
  • \[ v_{escape} = \sqrt{\frac{2GM}{R}} \]
Here:
  • \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \text{ m}^3\text{kg}^{-1}\text{s}^{-2} \)
  • \( M \) is the mass of the celestial body
  • \( R \) is the radius of the celestial body
The exercise explores the escape velocity from the sun, where:
  • \( M = 1.989 \times 10^{30} \text{ kg} \)
  • \( R = 6.96 \times 10^{8} \text{ m} \)
By substituting these values into our escape velocity formula, we can compute the speed necessary for particles to leave the Sun's gravitational field. This calculation is essential to understand whether hydrogen atoms could naturally leave the sun.
Hydrodynamics
Hydrodynamics is the study of fluids in motion, fundamentally tied to both physics principles and real-world applications. In thermal physics, particularly when discussing stellar objects like the sun, hydrodynamics helps explain how particles such as gases move and behave.
Thermal and gravitational influences cause gases on the sun to exhibit behaviors examined using hydrodynamic principles. This links closely with the RMS speed and escape velocity concepts. While RMS speed helps us understand the motion of particles at a micro level, hydrodynamics considers the fluid as a whole. It looks at large-scale behaviors impacted by forces, energy transfer, and dynamics.
  • Fluid motion is often turbulent and influenced by temperature gradients.
  • Pressure variations within the fluid are significant, especially in celestial environments.
  • Hydrodynamics tackles how these factors might cause some particles to reach speeds higher than average, potentially following phenomena like solar winds.
By integrating our understanding of RMS speeds and escape velocities with hydrodynamics, we provide a clearer picture of why, despite the high-energy environment of the sun, not all hydrogen atoms can escape its grasp. It’s the exceptional, high-speed hydrogen particles, possibly influenced by such dynamic processes, that may achieve the necessary escape velocity.

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Most popular questions from this chapter

(a) Compute the specific heat capacity at constant volume of nitrogen \(\left(\mathrm{N}_{2}\right)\) gas, and compare with the specific heat capacity of liquid water. The molar mass of \(\mathrm{N}_{2}\) is 28.0 \(\mathrm{g} / \mathrm{mol}\) . (b) You warm 1.00 \(\mathrm{kg}\) of water at a constant volume of 1.00 \(\mathrm{L}\) from \(20.0^{\circ} \mathrm{C}\) to \(30.0^{\circ} \mathrm{C}\) in a kettle. For the same amount of beat, how many kilograms of \(20.0^{\circ} \mathrm{C}\) air would you be able to warm to \(30.0^{\circ} \mathrm{C} ?\) What volume (in liters) would this air occupy at \(20.0^{\circ} \mathrm{C}\) and a pressure of 1.00 atm? Make the simplifying assumption that air is 100\(\% \mathrm{N}_{2}\) .

A diver observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the surface (where the pressure is 1.00 atm). The temperature at the bottom is \(4.0^{\circ} \mathrm{C},\) and the temperature at the surface is \(23.0^{\circ} \mathrm{C}\) , (a) What is the ratio of the volume of the bubble as it reaches the surface to its volume at the bottom? (b) Would it be safe for the diver to hold his breath while ascending from the bottom of the lake to the surface? Why or why not?

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.110 \(\mathrm{m}^{3}\) of air at pressure of 3.40 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.390 \(\mathrm{m}^{3} .\) If the temperature remains constant, what is the final value of the pressure?

The vapor pressure is the pressure of the vapor phase of a substance when it is in equilibrium with the solid or liquid phase of the substance. The relative humidity is the partial pressure of water vapor in the air divided by the vapor pressure of water at that same terperature, expressed as a percentage. The air is saturated when the humidity is 100\(\%\) . (a) The vapor pressure of water at \(20.0^{\circ} \mathrm{C}\) is \(2.34 \times 10^{3} \mathrm{Pa}\) . If the air temperature is \(20.0^{\circ} \mathrm{C}\) and the relative humidity is 60\(\%\) what is the partial pressure of water vapor in the atmosphere (that is, the pressure due to water vapor alone)? (b) Under the conditions of part (a), what is the mass of water in 1.00 \(\mathrm{m}^{3}\) of air? (The molar mass of water is 18.0 \(\mathrm{g} / \mathrm{mol}\) . Assume that water vapor can be treated as an ideal gas.)

During a test dive in \(1939,\) prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 \(\mathrm{m}\) . The temperature at the surface was \(27.0^{\circ} \mathrm{C},\) and at the bottom it was \(7.0^{\circ} \mathrm{C}\) . The density of seawater is 1030 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder 2.30 \(\mathrm{m}\) high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: You may ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell. \((b)\) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?

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