Question

Interstellar space far from any stars is usually filled with atomic hydrogen \((\mathrm{H})\) at a density of \(1.00 \mathrm{atom} / \mathrm{cm}^{3}\) and a very low temperature of \(2.73 \mathrm{~K}\) a) Determine the pressure in interstellar space. b) What is the root-mean-square speed of the atoms? c) What would be the edge length of a cube that would contain atoms with a total of 1.00 J of energy? 19.46 a) What is the root-mean-square speed for a collection of helium- 4 atoms at \(300 .\) K? b) What is the root-mean-square speed for a collection of helium-3 atoms at \(300 . \mathrm{K} ?\)

Short answer

What are the root-mean-square speeds for collections of helium-4 and helium-3 atoms at 300 K? Answer: The pressure of atomic hydrogen in interstellar space is approximately \(3.75 \times 10^{-23} \mathrm{Pa}\). The root-mean-square speed of the atoms is approximately \(542 \, \mathrm{m/s}\), and the edge length of a cube containing atoms with a total of 1.00 J of energy is approximately \(732 \, \mathrm{cm}\). The root-mean-square speeds for collections of helium-4 and helium-3 atoms at 300 K are approximately \(1360 \, \mathrm{m/s}\) and \(1570 \, \mathrm{m/s}\), respectively.

Step-by-step solution

Write the Ideal Gas Law

We must use the Ideal Gas Law equation to relate the pressure, volume, number of moles, and temperature of atomic hydrogen in interstellar space. The Ideal Gas Law is given by: $$PV=nRT$$

Calculate the number of moles

First, we need to convert the density of atomic hydrogen to moles per volume. We are given the density as \(1.00 \mathrm{atom} / \mathrm{cm}^{3}\). We can use Avogadro's number to convert atoms to moles: $$1.00 \mathrm{atom} \times \frac{1 \mathrm{mol}}{6.022 \times 10^{23} \mathrm{atoms}} = 1.66 \times 10^{-24} \mathrm{mol / cm^3}$$

Calculate the pressure

Now, we can substitute the values into the Ideal Gas Law equation and solve for pressure P: $$P=\frac{nRT}{V}=\frac{(1.66 \times 10^{-24} \mathrm{mol / cm^3})(8.314 \mathrm{J / mol \cdot K})(2.73 \mathrm{K})}{1 \mathrm{cm^3}} = 3.75 \times 10^{-23} \mathrm{Pa}$$ The pressure in interstellar space is approximately \(3.75 \times 10^{-23} \mathrm{Pa}\). ##Part b): What is the root-mean-square speed of the atoms?##

Write the formula for root-mean-square speed

Root-mean-square speed can be calculated using the following equation: $$v_{rms}=\sqrt{\frac{3RT}{M}}$$ where R is the gas constant, T is the temperature, and M is the molar mass of the gas.

Substitute values and calculate speed

The temperature T is given as \(2.73 \mathrm{K}\). The molar mass of hydrogen gas M is \(1.0079 \mathrm{g/mol} = 1.0079 \times 10^{-3} \mathrm{kg/mol}\). Substitute values into the equation: $$v_{rms} = \sqrt{\frac{3(8.314 \mathrm{J/mol \cdot K})(2.73 \mathrm{K})}{1.0079 \times 10^{-3} \mathrm{kg/mol}}} = 542 \mathrm{m/s}$$ The root-mean-square speed of the atoms is approximately \(542 \mathrm{m/s}\). ##Part c): What would be the edge length of a cube that would contain atoms with a total of 1.00 J of energy?##

Write the formula for kinetic energy

The kinetic energy of the atoms can be calculated using the equation: $$K=\frac{1}{2}mv^2$$

Calculate the mass of the atoms with 1 J of energy

Since the total energy is given as 1.00 J, we can set K = 1.00 J and solve for mass m: $$1 = \frac{1}{2}m(542 \mathrm{m/s})^2 \Rightarrow m = 6.78 \times 10^{-7} \mathrm{kg}$$

Calculate the number of moles

Now, we can convert the mass of hydrogen to moles using the molar mass: $$6.78 \times 10^{-7} \mathrm{kg} \times \frac{1 \mathrm{mol}}{1.0079 \times 10^{-3} \mathrm{kg}} = 6.73 \times 10^{-4} \mathrm{mol}$$

Calculate the edge length of the cube

We can use the Ideal Gas Law to find the volume of the cube: $$V=\frac{nRT}{P}=\frac{(6.73 \times 10^{-4} \mathrm{mol})(8.314 \mathrm{J/mol \cdot K})(2.73 \mathrm{K})}{3.75 \times 10^{-23} \mathrm{Pa}}=3.94\times10^8 \, \mathrm{cm^{3}}$$ Now find the edge length L of the cube: $$L = \sqrt[3]{V} = \sqrt[3]{3.94\times10^8 \, \mathrm{cm^{3}}} = 732 \, \mathrm{cm}$$ The edge length of the cube would be approximately \(732 \, \mathrm{cm}\). ##19.46 a): What is the root-mean-square speed for a collection of helium-4 atoms at \(300\mathrm{K}\)?##

Write the formula for root-mean-square speed

We can use the same equation for root-mean-square speed as before:$$v_{rms}=\sqrt{\frac{3RT}{M}}$$

Calculate the root-mean-square speed for helium-4 atoms

The molar mass of helium-4 is \(4.0026 \mathrm{g/mol} = 4.0026 \times 10^{-3} \mathrm{kg/mol}\). Substitute values and solve: $$v_{rms} = \sqrt{\frac{3(8.314 \mathrm{J/mol \cdot K})(300 \mathrm{K})}{4.0026 \times 10^{-3} \mathrm{kg/mol}}} = 1360 \mathrm{m/s}$$ The root-mean-square speed for a collection of helium-4 atoms at \(300 \mathrm{K}\) is approximately \(1360 \mathrm{m/s}\). ##19.46 b): What is the root-mean-square speed for a collection of helium-3 atoms at \(300 \mathrm{K}\)?##

Write the formula for root-mean-square speed

We can use the same equation for root-mean-square speed as before:$$v_{rms}=\sqrt{\frac{3RT}{M}}$$

Calculate the root-mean-square speed for helium-3 atoms

The molar mass of helium-3 is \(3.016 \mathrm{g/mol} = 3.016 \times 10^{-3} \mathrm{kg/mol}\). Substitute values and solve: $$v_{rms} = \sqrt{\frac{3(8.314 \mathrm{J/mol \cdot K})(300 \mathrm{K})}{3.016 \times 10^{-3} \mathrm{kg/mol}}} = 1570 \mathrm{m/s}$$ The root-mean-square speed for a collection of helium-3 atoms at \(300 \mathrm{K}\) is approximately \(1570 \mathrm{m/s}\).

Key concepts

Root-Mean-Square Speed

The root-mean-square (rms) speed is a way of representing the average speed of particles in a gas. It is derived from the kinetic theory of gases, which describes how gases behave on a molecular level. The rms speed is particularly important because it provides insight into the energy of the gas particles.

When we calculate the rms speed using the equation \(v_{rms} = \sqrt{\frac{3RT}{M}}\), we use the ideal gas constant (R), the temperature (T), and the molar mass of the gas (M). The square root indicates that we are taking the square root of the average of the particles' velocities squared. This is significant because it accounts for the fact that gas particles move in all directions and at various speeds.

For instance, in interstellar space, where the temperature is low, hydrogen atoms would have a relatively small rms speed, as we calculated in the example provided. Knowing the rms speed also allows us to understand other properties of the gas, such as its diffusion rate and how it might interact with its surroundings.

Interstellar Space Physics

Interstellar space, the vast voids between the stars, presents an extreme environment where the physics of gases can be observed in a setting much different from Earth's atmosphere. It's characterized by very low densities and temperatures, which makes it an intriguing case for the study of the ideal gas law in action.

In our exercise, we observed that even at a very low temperature of 2.73 K, which is close to the temperature of the cosmic microwave background radiation, we could calculate the pressure exerted by a sparse distribution of hydrogen atoms. This pressure, albeit extremely low, still adheres to the principles of the ideal gas law. The physics of interstellar space is also essential for informing our understanding of astrophysics, including the formation of stars, the behavior of nebulae, and the propagation of light through the cosmos.

Kinetic Theory of Gases

The kinetic theory of gases is a theoretical framework that explains the behavior of a hypothetical ideal gas. According to this theory, a gas is composed of a large number of small particles, all of which are in constant, random motion. The theory ties together macroscopic properties such as pressure, volume, and temperature with the microscopic actions of gas molecules.

It's the kinetic theory of gases that underlines the concept of rms speed, as it assumes that the particles are moving in all directions with a distribution of speeds. It also informs us that the temperature of a gas is a measure of the average kinetic energy of its particles. Applying this theory to problems in physics helps to evaluate more than just the speed of particles; it also facilitates a deeper understanding of energy transfer, phase changes, and reaction rates within gases.