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Interstellar space far from any stars is usually filled with atomic hydrogen (H) at a density of 1 atom/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 \(1.00 \mathrm{~J}\) of energy?

Short Answer

Expert verified
Answer: The pressure in interstellar space is approximately 3.76 x 10⁻¹⁷ Pa, the root-mean-square speed of the hydrogen atoms is approximately 617.58 m/s, and the edge length of a cube containing 1 J of energy is approximately 5.37 x 10⁷ cm.

Step by step solution

01

Find the pressure in interstellar space

To find the pressure in interstellar space, we will use the ideal gas law, \(PV = NkT\), where \(P\) is pressure, \(V\) is volume, \(N\) is the number of particles, \(k\) is Boltzmann's constant (\(1.38 \times 10^{-23} \mathrm{J/K}\)), and \(T\) is the temperature in Kelvin. Given the density \(\rho\) (in atoms per cm\(^3\)) and temperature \(T\), we can write the ideal gas law as: \(P = \rho kT\) Using the given values, \(\rho = 1 \mathrm{atom/cm^3}\) and \(T = 2.73 \mathrm{K}\): \(P = 1 \mathrm{atom/cm^3} \times 1.38 \times 10^{-23} \mathrm{J/K} \times 2.73 \mathrm{K}\)
02

Calculate the pressure

Now, we can calculate the pressure in interstellar space: \(P = 3.76 \times 10^{-23} \mathrm{J/cm^3}\) Since 1 Pa \(= 10^6 \mathrm{J/m^3}\), we can convert the pressure in J/cm\(^3\) to Pa: \(P = 3.76 \times 10^{-17} \mathrm{Pa}\) The pressure in interstellar space is approximately \(3.76 \times 10^{-17} \mathrm{Pa}\).
03

Determine the root-mean-square speed of the atoms

To find the root-mean-square speed (\(v_{rms}\)) of the hydrogen atoms, we can use the formula: \(v_{rms} = \sqrt{\frac{3kT}{m}}\) Where \(m\) is the mass of a hydrogen atom. Given that the atomic weight of hydrogen is 1, we can find the mass: \(m = 1 \times 1.67 \times 10^{-27} \mathrm{kg}\) Now, we can calculate the root-mean-square speed: \(v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \mathrm{J/K} \times 2.73 \mathrm{K}}{1.67 \times 10^{-27} \mathrm{kg}}}\)
04

Calculate the root-mean-square speed

Now, we can calculate the root-mean-square speed of the hydrogen atoms: \(v_{rms} \approx 617.58 \mathrm{m/s}\) The root-mean-square speed of the hydrogen atoms is approximately 617.58 m/s.
05

Determine the edge length of a cube containing 1 J of energy

To find the edge length of a cube containing 1 J of energy, we will use the formula for energy in an ideal gas: \(E = \frac{3}{2}NkT\) Where \(N\) is the number of atoms in the cube. Since we have \(E = 1 \mathrm{J}\), we can solve for \(N\): \(N = \frac{2E}{3kT} = \frac{2 \times 1 \mathrm{J}}{3 \times 1.38 \times 10^{-23} \mathrm{J/K} \times 2.73 \mathrm{K}}\) Now we can calculate the number of atoms in the cube: \(N \approx 1.54 \times 10^{23} \mathrm{atoms}\)
06

Calculate the edge length of the cube

Next, we use the density \(\rho = 1 \mathrm{atom/cm^3}\) to find the volume of the cube: \(V = \frac{N}{\rho} = \frac{1.54 \times 10^{23} \mathrm{atoms}}{1 \mathrm{atom/cm^3}} = 1.54 \times 10^{23} \mathrm{cm^3}\) To find the edge length \(a\) of a cube, we use the volume formula \(V=a^3\): \(a = \sqrt[3]{V} = \sqrt[3]{1.54 \times 10^{23} \mathrm{cm^3}}\) Now, we can calculate the edge length of the cube: \(a \approx 5.37 \times 10^7 \mathrm{cm}\) The edge length of a cube containing 1 J of energy is approximately \(5.37 \times 10^7 \mathrm{cm}\).

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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 key to understanding how gases behave under various conditions. In our exercise, we applied this law using the equation \( PV = NkT \). Here, \( P \) represents the pressure of the gas, \( V \) is the volume, \( N \) is the number of particles, and \( T \) represents the temperature. The constant \( k \) stands for Boltzmann's constant \( (1.38 \times 10^{-23} \mathrm{~J/K}) \).
This relation is powerful, providing a way to calculate unknown properties of a gas if some properties are known. In interstellar space, pressure is exceedingly small due to its low density, making the Ideal Gas Law essential in such calculations.
Atomic Hydrogen
Atomic hydrogen is the most abundant element in the universe and is a single proton often found in interstellar spaces. In our context, we have a density of 1 atom/cm\(^3\) in interstellar space. Understanding this helps us grasp the fundamental components of matter in space.
These hydrogen atoms are remote from stars and are at very low temperatures, like the 2.73 K given in the exercise. Even though sparse, their presence and properties like speed and pressure can be quantified using physics laws.
Root-Mean-Square Speed
The root-mean-square speed is a statistical measure of the speed of particles in a gas. It gives us the average speed of an atom in a substance and is calculated using \( v_{rms} = \sqrt{\frac{3kT}{m}} \).
Here, \( m \) is the mass of a hydrogen atom, which we found to be \( 1.67 \times 10^{-27} \mathrm{kg} \). This speed reflects the kinetic activity and energy within the gas.
In space, given the low temperatures, the speed is also lower around 617.58 m/s, providing insight into how slowly things move far from celestial bodies.
Pressure in Space
Pressure in interstellar space is notably low, around \(3.76 \times 10^{-17} \mathrm{Pa}\), due to the minimal presence of hydrogen atoms. Pressure, defined by force per unit area, arises here from atomic motion.
The sparse presence of particles and the low temperature result in a minuscule force acting over area, making calculations sensitive to precision. Despite the small values, pressure is vital in understanding forces that affect cosmic structures.
Energy in Ideal Gas
Energy in an ideal gas predominantly comes from the motion of its particles. This energy is given by \( E = \frac{3}{2}NkT \), with \( N \) indicating the number of atoms in the space.
For large energy, as in 1 Joule queried, the number of atoms increases dramatically. Our calculations revealed \( N \approx 1.54 \times 10^{23} \) atoms, showing how energy scales with particle numbers.
This insight helps in conceptualizing how vast spaces contain distributed energy, influencing everything from heat transfer to potential expansion fields in the cosmos.

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