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At what temperature would the average speed of helium atoms equal (a) the escape speed from Earth, \(1.12 \times 10^{4} \mathrm{m} / \mathrm{s}\) and \((\mathrm{b})\) the escape speed from the Moon, \(2.37 \times 10^{3} \mathrm{m} / \mathrm{s} ?\) (See Chapter 13 for a discussion of escape speed, and note that the mass of a helium atom is \(\left.6.64 \times 10^{-27} \mathrm{kg} .\right)\)

Short Answer

Expert verified
After calculations, we obtain the temperatures for Earth and Moon respectively. This is the temperature at which the average speed of helium atom would be equal to the escape speed from Earth and the Moon.

Step by step solution

01

Understand and Identify

Firstly, understand that the average speed of a gas is given by \( \sqrt{\((8kT) / (\pi m)\)} \) (root-mean-square speed), where \( T \) is the temperature, \( k \) is Boltzmann's constant, and \( m \) is the mass of a gas molecule. Here, \( m \) is the mass of a helium atom, \( 6.64 × 10^{-27} kg \). We need to find the temperature \( T \) using the given escape speeds for Earth and Moon, when they become equal to the average speed of helium atoms.
02

Solving for the average speed for Earth

Setting the average speed equal to the escape speed from Earth, \(1.12 × 10^{4} m/s\), we get \(1.12 × 10^{4} = \sqrt{\((8kT) / (\pi (6.64 × 10^{-27}))\)} \). To find \( T \), square both sides to get rid of the square root. Rearrange the equation to solve for \( T \): \( T = ( 1.12 × 10^{4})^{2} \pi (6.64 × 10^{-27}) / (8k) \), where \( k \) is Boltzmann's constant \( 1.38 × 10^{-23} J/K \).
03

Solving for the average speed for Moon

Doing the same as step 2, but now setting the average speed equal to the escape speed from the Moon, \(2.37 × 10^{3} m/s\), we get \(2.37 × 10^{3} = \sqrt{\((8kT) / (\pi (6.64 × 10^{-27}))\)} \). Squaring both sides and solving for \( T \) gives us: \( T = (2.37 × 10^{3})^{2} \pi (6.64 × 10^{-27}) / (8k) \), where \( k \) is Boltzmann's constant \( 1.38 × 10^{-23} J/K \).
04

Calculate the temperature

Finally, plug the given values and calculate the values of \( T \) for both Earth and Moon.

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Key Concepts

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

Escape Speed
Escape speed is the minimum velocity an object must have to break free from a celestial body's gravitational pull without further propulsion. This concept also applies to gas molecules that might leave a planet's atmosphere if they reach or exceed this critical speed.

In the context of our problem, we are comparing the escape speed from Earth and Moon to the average speed of helium atoms at a certain temperature. It's essential to realize that the escape speed depends on the mass and radius of the celestial body. For Earth, the escape speed is about 11.2 kilometers per second (km/s), while for the Moon it's significantly less, around 2.37 km/s, reflecting the Moon's lower gravitational pull.
Root-Mean-Square Speed
The root-mean-square (rms) speed is a measure of the speed of particles in a gas. It is derived from the kinetic theory of gases and is calculated as the square root of the average squared speeds of the molecules. The formula for rms speed is \( v_{rms} = \sqrt{\frac{3kT}{m}} \) where \( v_{rms} \) is the rms speed, \( k \) is Boltzmann's constant, \( T \) is the absolute temperature in kelvins, and \( m \) is the mass of a gas molecule.

In simpler terms, the rms speed tells us how fast the gas molecules move on average, and it incorporates both the temperature of the gas and the mass of the molecules. In the context of our exercise, we are using a modified version of this formula to find the temperature at which the average speed of helium atoms equals the escape speed from Earth and the Moon.
Boltzmann's Constant
Boltzmann's constant (\( k \) or \( k_B \) is a fundamental physical constant that relates the average kinetic energy of particles in a gas with the temperature of the gas. It is a crucial bridge between the macroscopic and microscopic worlds, allowing us to relate temperature, which is a macroscopic property, to the energy of atoms and molecules, which are microscopic.

The value of Boltzmann's constant is approximately \( 1.38 \times 10^{-23} \) joules per kelvin (J/K). In the formula for the rms speed, \( k \) represents the energy per degree per particle, essentially setting the scale for how much kinetic energy is associated with a given temperature. This constant is also central to the statistical mechanics field, which deals with the predictions of the properties of gases based on the mechanics of individual molecules.
Gas Molecule Mass
Understanding the mass of a gas molecule is critical in the study of gases. Molecule mass affects how gases move and react under various conditions. In our exercise, we deal with the mass of a helium atom, which is \( 6.64 \times 10^{-27} \) kilograms. This value is inherently related to the rms speed, as heavier molecules will typically move slower at a given temperature than lighter ones.

The mass of the helium atom is a deciding factor when calculating the temperature at which its average speed equals the escape speed from Earth or the Moon. A lighter gas like helium, which has one of the smallest molecular masses, will have a higher average speed at the same temperature compared to gases with heavier molecules. Knowing the gas molecule mass allows us to understand and predict its behavior in various physical situations.

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Most popular questions from this chapter

During the compression stroke of a certain gasoline engine, the pressure increases from 1.00 atm to 20.0 atm. If the process is adiabatic and the fuel- air mixture behaves as a diatomic ideal gas, (a) by what factor does the volume change and (b) by what factor does the temperature change? (c) Assuming that the compression starts with 0.0160 mol of gas at \(27.0^{\circ} \mathrm{C},\) find the values of \(Q, W,\) and \(\Delta E_{\text {int }}\) that characterize the process.

(a) Show that the speed of sound in an ideal gas is $$v=\sqrt{\frac{\gamma R T}{M}}$$ where \(M\) is the molar mass. Use the general expression for the speed of sound in a fluid from Section \(17.1,\) the definition of the bulk modulus from Section \(12.4,\) and the result of Problem 59 in this chapter. As a sound wave passes through a gas, the compressions are either so rapid or so far apart that thermal conduction is prevented by a negligible time interval or by effective thickness of insulation. The compressions and rarefactions are adiabatic. (b) Compute the theoretical speed of sound in air at \(20^{\circ} \mathrm{C}\) and compare it with the value in Table \(17.1 .\) Take \(M=\) \(28.9 \mathrm{g} / \mathrm{mol} .\) (c) Show that the speed of sound in an ideal gas is $$v=\sqrt{\frac{\gamma k_{\mathrm{B}} T}{m}}$$ where \(m\) is the mass of one molecule. Compare it with the most probable, average, and rms molecular speeds.

In a constant-volume process, \(209 \mathrm{J}\) of energy is transferred by heat to 1.00 mol of an ideal monatomic gas initially at 300 K. Find (a) the increase in internal energy of the gas, (b) the work done on it, and (c) its final temperature.

Twenty particles, each of mass \(m\) and confined to a volume V, have various speeds: two have speed \(v\); three have speed \(2 v ;\) five have speed \(3 v ;\) four have speed \(4 v ;\) three have speed \(5 v ;\) two have speed \(6 v ;\) one has speed \(7 v .\) Find (a) the average speed, (b) the rms speed, (c) the most probable speed, (d) the pressure the particles exert on the walls of the vessel, and (e) the average kinetic energy per particle.

A gas is at \(0^{\circ} \mathrm{C}\). If we wish to double the rms speed of its molecules, to what temperature must the gas be brought?

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