/*! 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 70 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 13.5 of Section 13.3 ). Use Appendix 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
After completing the calculations, the rms speed of the hydrogen atom and the escape speed from the sun can be determined. Compare the two values to determine if appreciable quantities of hydrogen can escape from the sun. Without the actual value calculations, we can't provide a concrete 'yes' or 'no' answer, but the steps above guide you on how to get to the answers.

Step by step solution

01

Calculate the rms speed

The rms (root mean square) speed of a gas particle is given by the equation \(v_{rms} = \sqrt{{3kT} \over m}\), where \(k\) is the Boltzmann constant (\(1.38 x 10^{-23} \mathrm{~J/K}\)), \(T\) is the temperature in Kelvin, and \(m\) is the mass of the particle. Substituting the appropriate values, we get \(v_{rms} = \sqrt{{3 \times 1.38 \times 10^{-23}} \times 5800 \over {1.67 \times 10^{-27}}}\)
02

Calculate the escape speed

The escape speed of a particle from a body of mass \(M\) and radius \(R\) is given by \(v_{esc} = \sqrt{2GM \over R}\), where \(G\) is the universal gravitational constant (\(6.67 \times 10^{-11} \mathrm{~m^3 kg^{-1} s^{-2}}\)), \(M\) is the mass of the sun (\(1.989 \times 10^{30} \mathrm{~kg}\)), and \(R\) is the radius of the sun (\(6.96 \times 10^{8} \mathrm{~m}\)). Plugging these into our equation, our escape speed comes out.
03

Determine Hydrogen Escape

Compare the calculated rms speed of the hydrogen atom to the escape speed. If the rms speed is equal or higher than the escape speed, then it means that a significant amount of hydrogen atoms can overcome the sun's gravitational pull and escape. If it lies far below the escape speed, it means that only a negligible quantity of hydrogen atoms possessing unusually high speed (in the tail of the Maxwell-Boltzmann distribution) can escape.

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Key Concepts

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

rms speed
The root mean square (rms) speed is a measure used in physics to express the average speed of particles in a gas. It accounts for the fact that particles move randomly and at different speeds. Importantly, the rms speed is derived from the kinetic theory of gases, which relates the speed of particles to temperature and mass. For a given gas at a specific temperature, the rms speed can be found using the formula: \[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \] where:
  • \( k \) is the Boltzmann constant, \( 1.38 \times 10^{-23} \mathrm{~J/K} \)
  • \( T \) is the absolute temperature measured in Kelvin
  • \( m \) is the mass of a single particle in kilograms
By substituting the specific values of hydrogen available at a temperature of 5800 K, we can compute the rms speed for hydrogen atoms. The higher the temperature, the faster the particles move, therefore increasing the rms speed.
escape velocity
Escape velocity is the speed that a particle or object must achieve to break free from a celestial body's gravitational pull entirely without any additional propulsion. For the sun, this concept is especially intriguing, given its massive gravitational influence. The escape speed from the sun's surface can be calculated using the formula: \[ v_{\text{esc}} = \sqrt{\frac{2GM}{R}} \] where:
  • \( G \) is the gravitational constant, \( 6.67 \times 10^{-11} \mathrm{~m^3 kg^{-1} s^{-2}} \)
  • \( M \) is the mass of the sun, approximately \( 1.989 \times 10^{30} \mathrm{~kg} \)
  • \( R \) is the radius of the sun, about \( 6.96 \times 10^{8} \mathrm{~m} \)
Understanding escape velocity helps us explore scenarios where gases, like hydrogen, might escape from celestial bodies like stars and planets. The vast gravitational pull of the sun implies that the escape velocity is quite high, meaning particles need a significant amount of energy to escape its grasp.
Maxwell-Boltzmann distribution
The Maxwell-Boltzmann distribution is a statistical means to describe the range of speeds that particles in a gas have at a given temperature. It is foundational in statistical mechanics. All particles in a gas don't move at the same speed, and this distribution helps describe the likelihood of finding a particle moving at a certain speed at a certain temperature. Key aspects of the Maxwell-Boltzmann distribution are as follows:
  • Most particles have speeds around a certain average value.
  • Few particles move extremely slowly or extremely quickly; these occur at the tails of the distribution curve.
  • The spread and peak of the distribution change with temperature—higher temperatures increase the average speed and broaden the distribution.
When considering if hydrogen can escape the sun, one must consider the distribution tail. Hydrogen atoms at the higher speeds (which are less common) could potentially reach or exceed the escape velocity, particularly if provided additional energy, despite the mean speed being lower.
gravitational constant
The gravitational constant, often denoted as \( G \), is a key factor in determining gravitational forces between masses. It is a fundamental constant in physics appearing in Newton's law of universal gravitation. The value of the gravitational constant is: \( 6.67 \times 10^{-11} \mathrm{~m^3 kg^{-1} s^{-2}} \). This constant helps calculate the forces between objects, illustrating how masses attract each other. Especially in contexts like celestial mechanics, it is used to compute escape velocities and study the orbits of planets and stars. Its small value reflects the weakness of the gravitational force compared to other fundamental forces.Understanding \( G \) is crucial for applications ranging from predicting the motions of planets to describing the escape velocities of objects from celestial bodies. It gives insight into how the vast distances and immense masses in space interact through gravity.

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Most popular questions from this chapter

The dark area in Fig. \(\mathbf{P} 18.83\) that appears devoid of stars is a dark nebula, a cold gas cloud in interstellar space that contains enough material to block out light from the stars behind it. A typical dark nebula is about 20 light-years in diameter and contains about 50 hydrogen atoms per cubic centimeter (monatomic hydrogen, not \(\mathrm{H}_{2}\) ) at about \(20 \mathrm{~K}\). (A lightyear is the distance light travels in vacuum in one year and is equal to \(\left.9.46 \times 10^{15} \mathrm{~m} .\right)\) (a) Estimate the mean free path for a hydrogen atom in a dark nebula. The radius of a hydrogen atom is \(5.0 \times 10^{-11} \mathrm{~m}\). (b) Estimate the rms speed of a hydrogen atom and the mean free time (the average time between collisions for a given atom). Based on this result, do you think that atomic collisions, such as those leading to \(\mathrm{H}_{2} \mathrm{~mol}-\) ecule formation, are very important in determining the composition of the nebula? (c) Estimate the pressure inside a dark nebula. (d) Compare the rms speed of a hydrogen atom to the escape speed at the surface of the nebula (assumed spherical). If the space around the nebula were a vacuum, would such a cloud be stable or would it tend to evaporate? (e) The stability of dark nebulae is explained by the presence of the interstellar medium (ISM), an even thinner gas that permeates space and in which the dark nebulae are embedded. Show that for dark nebulae to be in equilibrium with the ISM, the numbers of atoms per volume \((N / V)\) and the temperatures \((T)\) of dark nebulae and the ISM must be related by $$ \frac{(N / V)_{\text {nebula }}}{(N / V)_{\text {ISM }}}=\frac{T_{\text {ISM }}}{T_{\text {nebula }}} $$ (f) In the vicinity of the sun, the ISM contains about 1 hydrogen atom per \(200 \mathrm{~cm}^{3} .\) Estimate the temperature of the ISM in the vicinity of the sun. Compare to the temperature of the sun's surface, about \(5800 \mathrm{~K}\). Would a spacecraft coasting through interstellar space burn up? Why or why not?

How much heat does it take to increase the temperature of \(1.80 \mathrm{~mol}\) of an ideal gas by \(50.0 \mathrm{~K}\) near room temperature if the gas is held at constant volume and is (a) diatomic; (b) monatomic?

(a) What is the total translational kinetic energy of the air in an empty room that has dimensions \(8.00 \mathrm{~m} \times 12.00 \mathrm{~m} \times 4.00 \mathrm{~m}\) if the air is treated as an ideal gas at 1.00 atm? (b) What is the speed of a \(2000 \mathrm{~kg}\) automobile if its kinetic energy equals the translational kinetic energy calculated in part (a)?

A parcel of air over a campfire feels an upward buoyant force because the heated air is less dense than the surrounding air. By estimating the acceleration of the air immediately above a fire, one can estimate the fire's temperature. The mass of a volume \(V\) of air is \(n M_{\text {air }},\) where \(n\) is the number of moles of air molecules in the volume and \(M_{\text {air }}\) is the molar mass of air. The net upward force on a parcel of air above a fire is roughly given by \(\left(m_{\text {out }}-m_{\text {in }}\right) g,\) where \(m_{\text {out }}\) is the mass of a volume of ambient air and \(m_{\text {in }}\) is the mass of a similar volume of air in the hot zone. (a) Use the ideal-gas law, along with the knowledge that the pressure of the air above the fire is the same as that of the ambient air, to derive an expression for the acceleration \(a\) of an air parcel as a function of \(\left(T_{\text {out }} / T_{\text {in }}\right),\) where \(T_{\text {in }}\) is the absolute temperature of the air above the fire and \(T_{\text {out }}\) is the absolute temperature of the ambient air. (b) Rearrange your formula from part (a) to obtain an expression for \(T_{\text {in }}\) as a function of \(T_{\text {out }}\) and \(a\). (c) Based on your experience with campfires, estimate the acceleration of the air above the fire by comparing in your mind the upward trajectory of sparks with the acceleration of falling objects. Thus you can estimate \(a\) as a multiple of \(g .\) (d) Assuming an ambient temperature of \(15^{\circ} \mathrm{C}\), use your formula and your estimate of \(a\) to estimate the temperature of the fire.

(a) A deuteron, \({ }_{1}^{2} \mathrm{H},\) is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a nuclear fusion reactor must be heated to about 300 million \(\mathrm{K}\). What is the rms speed of the deuterons? Is this a significant fraction of the speed of light in vacuum \(\left(c=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\right) ?(\mathrm{~b})\) What would the temperature of the plasma be if the deuterons had an rms speed equal to \(0.10 \mathrm{c}\) ?

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