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Chapter 19: Problem 26

What is the average translational kinetic energy of nitrogen molecules at \(1600 \mathrm{~K}\) ?

Short Answer

Expert verified
The average translational kinetic energy is \(3.312 \times 10^{-20} \text{ J}\).

Step by step solution

01

Identify the Formula for Average Translational Kinetic Energy

The average translational kinetic energy of a molecule can be determined using the formula \[ \bar{E}_k = \frac{3}{2} k_B T \]where \(\bar{E}_k\) is the average translational kinetic energy, \(k_B\) is the Boltzmann constant, and \(T\) is the temperature in Kelvin (K).
02

Substitute Known Values into the Formula

The given temperature is \( T = 1600 \text{ K} \), and the Boltzmann constant is \( k_B = 1.38 \times 10^{-23} \text{ J/K} \). Substitute these values into the formula:\[ \bar{E}_k = \frac{3}{2} \times 1.38 \times 10^{-23} \times 1600 \]
03

Calculate the Average Translational Kinetic Energy

Perform the calculations:\[ \bar{E}_k = \frac{3}{2} \times 1.38 \times 10^{-23} \times 1600 = 3.312 \times 10^{-20} \text{ J} \]Thus, the average translational kinetic energy of nitrogen molecules at \(1600 \text{ K}\) is \(3.312 \times 10^{-20} \text{ J}\).

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Key Concepts

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

Boltzmann Constant
The Boltzmann constant
  • Fundamental physical constant denoted by \( k_B \).
  • Acts as a bridge between macroscopic and microscopic physics.
  • Value: \( 1.38 \times 10^{-23} \text{ J/K} \).
This small number plays a colossal role in connecting temperature with energy at the particle level. It helps translate temperature into kinetic energy, which in turn implies how fast particles move. Essentially, whenever you deal with temperature at the molecular scale, the Boltzmann constant helps quantify the energy involved.
The constant is crucial in statistical mechanics, governing the distribution of particles in thermal equilibrium. It demonstrates that as temperature rises, molecules move faster, increasing their kinetic energy.
Average Kinetic Energy
Average kinetic energy
  • Relates to the motion of particles like atoms and molecules.
  • Expressed as \( \bar{E}_k = \frac{3}{2} k_B T \).
In this formula, \( \bar{E}_k \) represents the average kinetic energy of a particle, linking directly to temperature \( T \) through the Boltzmann constant \( k_B \). This means the hotter the substance, the greater its particles move, increasing their kinetic energy.
Average kinetic energy doesn't just inform us about speed. It's fundamental in understanding the gas behaviors and impacts the laws governing them.
The simplicity of the formula shows how a measure like Kelvin temperature can give a direct insight into molecular speed. Additionally, it's important in calculating energies in systems behaving ideally.
Ideal Gas Law
The Ideal Gas Law
  • Equation: \( PV = nRT \)
  • Describes behavior of an "ideal" gas.
  • Variables: Pressure (P), Volume (V), amount of substance (n), ideal gas constant (R), and Temperature (T).
This law is a pivotal framework for understanding gas behaviors under various conditions. Although theoretical, it serves as an excellent approximation for real-world gases at higher temperatures and low pressures.
In terms of kinetic energy, it correlates volume and pressure to temperature, indirectly inferring how these factors affect particle motion and energy. The law shows that if you know four of the variables, you can predict the fifth. This makes it a versatile tool in physics and chemistry.
Besides, it gives insights on how gases respond to changes, ideal for understanding everything from weather patterns to the working principles of engines.

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Most popular questions from this chapter

An ideal gas with \(3.00 \mathrm{~mol}\) is initially in state 1 with pressure \(p_{1}=20.0\) 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 \(3 ?\)

At \(20^{\circ} \mathrm{C}\) and 750 torr pressure, the mean free paths for argon gas (Ar) and nitrogen gas \(\left(\mathrm{N}_{2}\right)\) are \(\lambda_{\mathrm{Ar}}=9.9 \times 10^{-6} \mathrm{~cm}\) and \(\lambda_{\mathrm{N}_{2}}=\) \(27.5 \times 10^{-6} \mathrm{~cm} .\) (a) Find the ratio of the diameter of an \(\mathrm{Ar}\) atom to that of an \(\mathrm{N}_{2}\) molecule. What is the mean free path of argon at (b) \(20^{\circ} \mathrm{C}\) and 150 torr, and \((\mathrm{c})-40^{\circ} \mathrm{C}\) and 750 torr?

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 automobile tire has a volume of \(1.64 \times 10^{-2} \mathrm{~m}^{3}\) and contains air at a gauge pressure (pressure above atmospheric pressure) of \(165 \mathrm{kPa}\) when the temperature is \(0.00^{\circ} \mathrm{C}\). What is the gauge pressure of the air in the tires when its temperature rises to \(27.0^{\circ} \mathrm{C}\) and its volume increases to \(1.67 \times 10^{-2} \mathrm{~m}^{3} ?\) Assume atmospheric pressure is \(1.01 \times 10^{5} \mathrm{~Pa}\).

(a) What is the volume occupied by \(1.00\) mol of an ideal gas at standard conditions \(-\) that is, \(1.00\) atm \(\left(=1.01 \times 10^{5} \mathrm{~Pa}\right)\) and \(273 \mathrm{~K} ?\) (b) Show that the number of molecules per cubic centimeter (the Loschmidt number) at standard conditions is \(2.69 \times 10^{9}\).
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