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Two moles of an ideal gas are placed in a container whose volume is \(8.5 \times 10^{-3} \mathrm{m}^{3} .\) The absolute pressure of the gas is \(4.5 \times 10^{5} \mathrm{Pa} .\) What is the average translational kinetic energy of a molecule of the gas?

Short Answer

Expert verified
The average translational kinetic energy is approximately \(4.76 \times 10^{-21} \text{J}.\)

Step by step solution

01

Determine the Temperature of the Gas

To find the average translational kinetic energy of a molecule, we first need to find the temperature \( T \) of the gas. We can use the Ideal Gas Law, given by \( PV = nRT \), where \( P \) is the pressure \( (4.5 \times 10^5\, \text{Pa}) \), \( V \) is the volume \( (8.5 \times 10^{-3} \text{m}^3) \), \( n \) is the number of moles \( (2) \), and \( R \) is the ideal gas constant \( (8.314 \text{J/mol K}) \). Rearrange for \( T \): \[ T = \frac{PV}{nR} = \frac{4.5 \times 10^5 \times 8.5 \times 10^{-3}}{2 \times 8.314} \approx 230 \text{ K}. \]
02

Use the Formula for Average Translational Kinetic Energy

The average translational kinetic energy \( (KE) \) of a molecule in an ideal gas is given by the expression: \[ KE = \frac{3}{2} kT, \] where \( k \) is Boltzmann's constant \( (1.38 \times 10^{-23} \text{J/K}) \). Substitute the temperature we found: \[ KE = \frac{3}{2} \times 1.38 \times 10^{-23} \times 230 \approx 4.76 \times 10^{-21} \text{J}. \]

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Key Concepts

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

Average Translational Kinetic Energy
The average translational kinetic energy is an important concept in understanding how particles behave in an ideal gas. It provides insight into the energy associated with the motion of individual gas molecules.
To calculate this energy for a molecule in an ideal gas, you can use the formula:
  • KE = \(\frac{3}{2} kT\)
Here.
- KE stands for the average translational kinetic energy.
- \(k\) is Boltzmann's constant, \(1.38 \times 10^{-23} \text{J/K}\). This constant essentially acts as a bridge between macroscopic and microscopic worlds.
- \(T\) is the absolute temperature in Kelvin.
This relationship tells us that as the temperature of the gas increases, so does the average kinetic energy of its molecules.
It's crucial in understanding that energy and temperature are directly linked at the molecular level.
Temperature Determination
Knowing the temperature is vital in many physics calculations involving gases. In our problem, we determine it using the Ideal Gas Law, which is stated as \(PV = nRT\).
Here's a breakdown of the terms:
  • \(P\) is pressure, given as \(4.5 \times 10^5 \, \text{Pa}\), which means Pascals.
  • \(V\) is volume, provided as \(8.5 \times 10^{-3} \, \text{m}^3\).
  • \(n\) is the number of moles, \(2\) here.
  • \(R\) is the ideal gas constant, \(8.314 \, \text{J/mol K}\), which we will discuss in detail below.
To find the temperature \(T\), rearrange the formula to: \(T = \frac{PV}{nR}\). Substitute in the values to get \(T \approx 230 \, \text{K}\).
This temperature calculation helps in determining the kinetic energy of particles, showing the strong connection between these concepts.
Ideal Gas Constant
The ideal gas constant \(R\) is pivotal in linking many properties of gases. It appears in the Ideal Gas Law \(PV = nRT\) and has a value of \(8.314 \, \text{J/mol K}\). This constant enables physical relationships between pressure, volume, temperature, and number of moles of a gas.
  • Unit Explanation: \(\text{J/mol K}\) represents Joules per mole per Kelvin. These units are crucial because they directly relate energy (Joules) to temperature (Kelvin) and amount of substance (moles).
  • Application: \(R\) is used in situations involving thermodynamic processes and calculations, highlighting its importance in fields like chemistry and physics.
Grasping \(R\) helps understand how energy changes within a system as gases undergo different transformations. This concept is not only theoretical but also applicable in practical scenarios involving gas behaviors.

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Most popular questions from this chapter

A tube has a length of \(0.015 \mathrm{m}\) and a cross-sectional area of \(7.0 \times \mathrm{x}\) \(10^{-4} \mathrm{m}^{2} .\) The tube is filled with a solution of sucrose in water. The diffusion constant of sucrose in water is \(5.0 \times 10^{-10} \mathrm{m}^{2} / \mathrm{s} .\) A difference in concentration of \(3.0 \times 10^{-3} \mathrm{kg} / \mathrm{m}^{3}\) is maintained between the ends of the tube. How much time is required for \(8.0 \times 10^{-13} \mathrm{kg}\) of sucrose to be transported through the tube?

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