/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A container holds 2.0 mol of gas... [FREE SOLUTION] | 91Ó°ÊÓ

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A container holds 2.0 mol of gas. The total average kinetic energy of the gas molecules in the container is equal to the kinetic energy of an \(8.0 \times 10^{-3}-\mathrm{kg}\) bullet with a speed of \(770 \mathrm{m} / \mathrm{s} .\) What is the Kelvin temperature of the gas?

Short Answer

Expert verified
The Kelvin temperature of the gas is approximately 95.08 K.

Step by step solution

01

Calculate the bullet's kinetic energy

The kinetic energy \( KE \) of an object is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity. Here, \( m = 8.0 \times 10^{-3} \) kg and \( v = 770 \) m/s. Substitute these values into the formula:\[KE = \frac{1}{2} \times 8.0 \times 10^{-3} \times (770)^2\]Calculate this to find the kinetic energy of the bullet.
02

Solve for the bullet's kinetic energy

Plug the numbers into the kinetic energy formula:\[KE = \frac{1}{2} \times 8.0 \times 10^{-3} \times 592900\]\[KE = 0.004 \times 592900\]\[KE = 2371.6 \text{ J}\]The kinetic energy of the bullet is 2371.6 Joules.
03

Relate bullet's kinetic energy to gas molecules

The total average kinetic energy of the gas molecules is given to be the same as the bullet's kinetic energy, 2371.6 J. Use this to find the temperature of the gas in Kelvin.
04

Use the kinetic energy formula for gases

The average kinetic energy of \( n \) moles of a gas is given by \( \text{Total KE} = \frac{3}{2} nRT \), where \( R = 8.314 \text{ J/(mol K)} \) is the ideal gas constant and \( T \) is the temperature in Kelvin. Equate this to the bullet's kinetic energy:\[\frac{3}{2} \times 2.0 \times 8.314 \times T = 2371.6\]Solve for \( T \).
05

Solve for the Kelvin temperature

Rearrange the equation to solve for \( T \):\[3 \times 8.314 \times T = 2371.6\]\[24.942 \times T = 2371.6\]\[T = \frac{2371.6}{24.942}\]\[T \approx 95.08 \text{ K}\]The temperature of the gas is approximately 95.08 Kelvin.

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Key Concepts

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

Moles of Gas
When dealing with gases, understanding the concept of moles is crucial. A mole is a unit of measurement used in chemistry to express amounts of a chemical substance. One mole (mol) represents Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles, be they atoms, molecules, ions, or electrons.

This massive number helps chemists to quantify substances in a manageable way since atoms and molecules are exceedingly small. In the context of gases, moles help us express the number of gas molecules present in a container without needing to count each particle individually.

In our exercise, the container has 2.0 moles of gas, indicating the quantity of gas molecules involved. This is significant because it relates directly to how we calculate properties like average kinetic energy, which depends on both the quantity of gas and temperature.
Ideal Gas Constant
The ideal gas constant, denoted by \(R\), plays a fundamental role in the equations related to gases. It is a constant that appears in the ideal gas law and is essential for calculations involving gases under ideal conditions. The value of \(R\) is \(8.314 \text{ J/(mol K)}\), which bridges the gap between energy, temperature, and moles in these calculations.

This constant helps us in relating various physical properties of gases, such as pressure, volume, temperature, and amount in moles. It arises from the combination of the Boltzmann constant with Avogadro's number, making it crucial for converting many aspects of molecular calculations into lab-scale realities.

In the exercise scenario, using the ideal gas constant allows us to relate the kinetic energy of a given number of moles of gas to its temperature, which eventually helps us derive the temperature in Kelvin.
Kelvin Temperature
Temperature is a measure of the average kinetic energy of the particles in a substance. In scientific contexts, particularly in gas calculations, the Kelvin scale is the standard unit of temperature used.

Unlike Celsius, Kelvin starts at absolute zero, where particle motion theoretically stops. Therefore, 0 Kelvin represents the lowest possible energy state of matter. This scale helps simplify equations, as it directly relates temperature to kinetic energy, which is not the case with Celsius or Fahrenheit.

In our exercise, the temperature determined for the gas is approximately 95.08 Kelvin, showing that the calculation bridges molecular motion and thermal energy in a straightforward way. The conversion to Kelvin ensures precision in scientific calculations, allowing for consistency when comparing different systems.
Average Kinetic Energy
Average kinetic energy is a key concept in understanding gases. It refers to the average energy of motion of the gas molecules within a given container. This energy is directly related to the temperature of the gas in question.

The fundamental relation is that higher temperature means that molecules move more quickly, resulting in greater kinetic energy. For an ideal gas, the average kinetic energy per molecule can be expressed as \(E_{avg} = \frac{3}{2} kT\), where \(k\) is the Boltzmann constant. However, for moles of gas, it's given by \(E_{avg} = \frac{3}{2} nRT\) where \(n\) is the number of moles and \(R\) is the ideal gas constant.

In the original exercise, the given moles of gas have an average kinetic energy matching that of a speeding bullet, which connects molecular motion to real-world kinetic energy, offering an intuitive understanding of molecular dynamics.
Gas Laws
Gas laws describe the behavior of gases and how they interact with factors like pressure, volume, and temperature. These relationships are crucial for understanding and predicting how gases will behave under various conditions.

The most well-known of these laws is the ideal gas law, described by the equation \(PV = nRT\), which shows how pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) of a gas relate to the number of moles \(n\) and the ideal gas constant \(R\).

In practice, these laws help chemists and scientists perform calculations and create models that predict gas behavior accurately, crucial in fields ranging from meteorology to chemical engineering.

In the exercise, these gas laws and their principles underpin the calculations, allowing us to connect the theoretical kinetic energy of a gas with its measurable thermal energy, revealing the real-world applications of these foundational scientific concepts.

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Most popular questions from this chapter

The average value of the squared speed \(\overline{v^{2}}\) does not equal the square of the average speed \((\bar{v})^{2} .\) To verify this fact, consider three particles with the following speeds: \(v_{1}=3.0 \mathrm{m} / \mathrm{s}, v_{2}=7.0 \mathrm{m} / \mathrm{s},\) and \(\quad v_{3}=9.0 \mathrm{m} / \mathrm{s} . \quad\) Calculate (a) \(\quad \overline{v^{2}}=\frac{1}{3}\left(v_{1}^{2}+v_{2}^{2}+v_{3}^{2}\right) \quad\) and (b) \((\bar{v})^{2}=\left[\frac{1}{3}\left(v_{1}+v_{2}+v_{3}\right)\right]^{2}\)

What is the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of nitrogen gas (molecular mass \(=28 \mathrm{u})\) at a pressure of 2.0 atmospheres and a temperature of \(310 \mathrm{K} ?\)

Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomic gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio \(\frac{\overline{\mathrm{KE}}_{\mathrm{Krypton}}}{\overline{\mathrm{KE}}_{\mathrm{Argon}}}\) of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom.

A \(0.030-\mathrm{m}^{3}\) container is initially evacuated. Then, \(4.0 \mathrm{g}\) of water is placed in the container, and, after some time, all the water evaporates. If the temperature of the water vapor is \(388 \mathrm{K},\) what is its pressure?

A tube has a length of \(0.015 \mathrm{m}\) and a cross-sectional area of \(7.0 \times\) \(10^{-4} \mathrm{m}^{2} .\) 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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