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Chapter 19: Problem 41

A hydrogen molecule (diameter \(\left.1.0 \times 10^{-8} \mathrm{~cm}\right)\), traveling at the rms speed, escapes from a \(4000 \mathrm{~K}\) furnace into a chamber containing cold argon atoms (diameter \(3.0 \times 10^{-8} \mathrm{~cm}\) ) at a density of \(4.0 \times 10^{19}\) atoms \(/ \mathrm{cm}^{3}\). (a) What is the speed of the hydrogen molecule? (b) If it collides with an argon atom, what is the closest their centers can be, considering each as spherical? (c) What is the initial number of collisions per second experienced by the hydrogen molecule? (Hint: Assume that the argon atoms are stationary. Then the mean free path of the hydrogen molecule is given by Eq. \(19-26\) and not Eq. \(19-25 .)\)

Short Answer

Expert verified
(a) 4060 m/s, (b) 2.0 x 10^-8 cm, (c) 1.02 x 10^6 collisions/s.

Step by step solution

01

Calculate RMS Speed

The root mean square (RMS) speed of a hydrogen molecule can be calculated using the formula for the RMS speed of gas particles: \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is Boltzmann's constant \( (1.38 \times 10^{-23} \text{ J/K}) \), \( T \) is the temperature \( (4000 \text{ K}) \), and \( m \) is the mass of a hydrogen molecule \( (2 \times 1.67 \times 10^{-27} \text{ kg}) \). Substituting these values, we find:\[v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 4000}{2 \times 1.67 \times 10^{-27}}}\]Calculating this gives \( v_{rms} = \sqrt{1.656 \times 10^{5}} \approx 4060 \text{ m/s} \).
02

Determine Closest Approach

When a hydrogen molecule collides with an argon atom, we consider both as spheres. The closest the centers can approach is the sum of their radii. Given the diameters, the radius of hydrogen is \( 0.5 \times 10^{-8} \text{ cm} \) and of argon is \( 1.5 \times 10^{-8} \text{ cm} \). Therefore, the closest distance is:\[0.5 \times 10^{-8} + 1.5 \times 10^{-8} = 2.0 \times 10^{-8} \text{ cm}\]
03

Calculate Collision Frequency

The initial number of collisions per second (collision frequency) is determined from the mean free path \( \lambda \), which can be expressed as \( \lambda = \frac{kT}{\sqrt{2} \pi d^2 n} \), where \( d \) is the effective diameter \( 2.0 \times 10^{-8} \text{ cm} \) from Step 2, and \( n \) is the number density of argon \( (4.0 \times 10^{19} \text{ cm}^{-3}) \).The mean free path is:\[\lambda = \frac{1}{\sqrt{2} \pi \left(2.0 \times 10^{-8}\right)^2 \times 4.0 \times 10^{19} \text{ cm}^{-3}}\]Calculating this yield \( \lambda \approx 3.98 \times 10^{-3} \text{ cm} \).Finally, the collision frequency \( f \) is \( f = \frac{v_{rms}}{\lambda} \):\[f = \frac{4060}{3.98 \times 10^{-3}} \approx 1.02 \times 10^{6} \text{ collisions/s}\]

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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 crucial concept in the kinetic theory of gases. It represents the speed of particles in a gas, taking into account their mass and the temperature of the system.

For a hydrogen molecule at a high temperature like 4000 K, the RMS speed can be calculated using
  • the formula: \( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \).
  • Here, \( k \) is Boltzmann's constant (\(1.38 \times 10^{-23} \text{ J/K}\)), and \( m \) is the mass of a hydrogen molecule.
By plugging in the values (temperature \( T = 4000 \text{ K} \) and the mass of a hydrogen molecule \( m = 2 \times 1.67 \times 10^{-27} \text{ kg} \)), we can calculate the RMS speed to find that the hydrogen molecule moves at approximately 4060 m/s.

This high speed is typical for small, light molecules at such high temperatures, illustrating how temperature influences particle motion in gases.
Mean free path
Mean free path is a measure of the average distance a molecule travels before colliding with another particle in a gas.

In the context of the exercise, when a hydrogen molecule moves through argon gas, the mean free path determines how far the molecule can go without a collision. It is calculated using:
  • \( \lambda = \frac{1}{\sqrt{2} \pi d^2 n} \).
Here, \( d \) represents the effective diameter from both gas particles (in this case derived as 2.0 \(\times 10^{-8} \text{ cm}\)), and \( n \) is the number density of argon atoms (\(4.0 \times 10^{19} \text{ cm}^{-3}\)).

After performing the calculations, we find the mean free path \( \lambda \) to be about 3.98 \( \times 10^{-3} \text{ cm}\). This small mean free path indicates frequent collisions, which is expected when a fast hydrogen molecule traverses dense argon atoms.
Collision frequency
Understanding collision frequency helps gauge how often molecules in a gas collide, impacting gas properties like pressure and diffusion.

The collision frequency \( f \) is derived from both RMS speed and mean free path, as shown by the equation:
  • \( f = \frac{v_{\text{rms}}}{\lambda} \).
In this exercise, using the derived RMS speed of 4060 m/s and mean free path of 3.98 \( \times 10^{-3} \text{ cm} \), we calculate that the hydrogen molecule experiences approximately 1.02 \( \times 10^6 \) collisions per second.

Such a high collision rate is typical in dense gases, where the continuous exchange of thermal energy between molecules rapidly alters their speeds and directions.

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Most popular questions from this chapter

Gold has a molar mass of \(197 \mathrm{~g} / \mathrm{mol}\). (a) How many moles of gold are in a \(2.50 \mathrm{~g}\) sample of pure gold? (b) How many atoms are in the sample?

An ideal gas with \(3.00 \mathrm{~mol}\) is initially in state 1 with pressure \(p_{1}=20.0 \mathrm{~atm}\) and volume \(V_{1}=1500 \mathrm{~cm}^{3} .\) First it is taken to state 2 with pressure \(p_{2}=1.50 p_{1}\) and volume \(V_{2}=2.00 V_{1} .\) Then it is taken to state 3 with pressure \(p_{3}=2.00 p_{1}\) and volume \(V_{3}=0.500 V_{1} .\) What is the temperature of the gas in (a) state 1 and (b) state \(2 ?\) (c) What is the net change in internal energy from state 1 to state \(3 ?\)

Ten particles are moving with the following speeds: four at \(200 \mathrm{~m} / \mathrm{s},\) two at \(500 \mathrm{~m} / \mathrm{s},\) and four at \(600 \mathrm{~m} / \mathrm{s} .\) Calculate their (a) average and (b) rms speeds. (c) Is \(v_{\text {rms }}>v_{\text {avg }} ?\)

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Two containers are at the same temperature. The gas in the first container is at pressure \(p_{1}\) and has molecules with mass \(m_{1}\) and root-mean-square speed \(v_{\text {rmsl }}\). The gas in the second is at pressure \(2 p_{1}\) and has molecules with mass \(m_{2}\) and average speed \(v_{\text {avg } 2}=2 v_{\text {rms } 1}\). Find the ratio \(m_{1} / m_{2}\) of the masses of their molecules.
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