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A vessel containing one mole of a monatomic ideal gas (molecular weight \(=20 \mathrm{~g} \mathrm{~mol}^{-1}\) ) is moving on a floor at a speed of \(50 \mathrm{~m} \mathrm{~s}^{-1}\). The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.

Short Answer

Expert verified
The rise in temperature is approximately 2.004 K.

Step by step solution

01

Calculate Initial Kinetic Energy of the Vessel

The initial kinetic energy of the vessel can be calculated using the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the gas and \( v \) is the velocity. Given the molecular weight \( 20 \text{ g/mol} \), convert it to kilograms to get \( 0.02 \text{ kg/mol} \). Since there is one mole of gas, the mass \( m = 0.02 \text{ kg} \), and \( v = 50 \text{ m/s} \). Thus, \( KE = \frac{1}{2} \times 0.02 \times 50^2 = 25 \text{ J} \).
02

Relate Mechanical Energy to Internal Energy

Since the vessel is stopped suddenly, the entire initial kinetic energy \( 25 \text{ J} \) is transferred to the internal energy of the gas. For a monatomic ideal gas, the increase in internal energy is related to the rise in temperature \( \Delta T \) by the equation \( \Delta U = \frac{3}{2} nR\Delta T \), where \( n = 1 \) mol and \( R = 8.314 \text{ J/mol K} \).
03

Solve for Temperature Increase

Using the relation from Step 2, equate the change in internal energy to the absorbed energy: \( 25 = \frac{3}{2} \times 1 \times 8.314 \times \Delta T \). Solving for \( \Delta T \), we get \[ 25 = 12.471 \times \Delta T \] \[ \Delta T = \frac{25}{12.471} \approx 2.004 \text{ K} \].

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Key Concepts

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

Kinetic Energy
Kinetic energy is an essential concept to understand when analyzing the motion of objects. In physics, kinetic energy refers to the energy an object has due to its motion. For any object moving at a certain velocity, you can calculate its kinetic energy using the formula \( KE = \frac{1}{2} mv^2 \). This equation states that kinetic energy depends on the mass \( m \) of the object and its velocity \( v \).
In the context of a monatomic ideal gas, as described in the exercise, the kinetic energy of the entire vessel containing the gas is considered. It's important to note that as the vessel comes to a sudden stop, the kinetic energy initially associated with the motion of the vessel is lost and has to be accounted for elsewhere, in this scenario by converting to the internal energy of the gas.
  • Mass and velocity play a crucial role in determining kinetic energy.
  • When a moving object stops, its kinetic energy is transformed.
Understanding kinetic energy helps in analyzing and predicting how energy is redistributed within a system.
Internal Energy
Internal energy in a system refers to the total energy contained within it, encompassing both kinetic and potential energy at the molecular level. For gases, especially ideal monatomic gases like in the given exercise, internal energy is chiefly due to the kinetic energy of the particles.
When a vessel containing a monatomic ideal gas stops suddenly, the kinetic energy lost from the vessel's motion is transferred to the internal energy of the gas. This transformation leads to changes in other properties of the gas, such as temperature.
The internal energy change, \( \Delta U \), can be associated with temperature change using the formula \( \Delta U = \frac{3}{2} nR\Delta T \), where \( n \) is the number of moles, and \( R \) is the gas constant. This clearly shows that internal energy changes are directly related to how temperature varies based on absorbed heat or mechanical energy shuffles within the system.
  • Internal energy comprises different forms of energy within particles.
  • The conversion of mechanical to internal energy affects temperature.
This interplay helps understand how energy moves within substances and impacts their physical states.
Temperature Increase
The rise in temperature is a direct consequence of energy transformations within a system. In the context of gases, when energy is added, whether through heat or mechanical means, such as the sudden stop of a moving vessel, this energy increases the internal energy of the gas, causing a temperature increase.
For a monatomic ideal gas, we focus on the relationship between temperature changes and internal energy changes. Using the formula \( \Delta T = \frac{\Delta U}{\frac{3}{2} nR} \), we can compute how much the temperature rises when a specific amount of internal energy change occurs. In the exercise, the vessel's stop leads to an internal energy increase equal to the kinetic energy, resulting in a temperature increase.
  • Energy transformations result in temperature changes.
  • Temperature increase is linked to how a system absorbs energy.
Grasping the concept of temperature rise in gases helps in understanding broader phenomena, like why gases expand when heated or the principles behind engines and refrigerators.

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Most popular questions from this chapter

Two samples \(A\) and \(B\) of the same gas have equal volumes and pressures. The gas in sample \(A\) is expanded isothermally to double its volume and the gas in \(B\) is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that \(\gamma\) satisfies the equation \(1-2^{1-\gamma}=(\gamma-1) \ln 2 .\)

A sample of air weighing \(1.18 \mathrm{~g}\) occupies \(1 \cdot 0 \times 10^{3} \mathrm{~cm}^{3}\) when kept at \(300 \mathrm{~K}\) and \(1-0 \times 10^{5} \mathrm{~Pa}\). When \(2 \cdot 0 \mathrm{cal}\) of heat is added to it at constant volume, its temperature increases by \(1^{\circ} \mathrm{C} .\) Calculate the amount of heat needed to increase the temperature of air by \(1^{\circ} \mathrm{C}\) at constant pressure if the mechanical equivalent of heat is \(4.2 \times 10^{-7}\) erg \(\mathrm{cal}^{-1}\). Assume that air behaves as an ideal gas.

A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can and the piston are adiabatic. The initial pressure, volume and temperature of the gas are \(100 \mathrm{kPa}, 400 \mathrm{~cm}^{3}\) and \(300 \mathrm{~K}\) respectively. The ratio of the specific heat capacities of the gas is \(C_{p} / C_{V}=1 \cdot 5\). Find the pressure and the temperature of the gas if it is (a) suddenly compressed (b) slowly compressed to \(100 \mathrm{~cm}^{3}\)

The volume of an ideal gas \((\gamma=1 \cdot 5)\) is changed adiabatically from \(4.00\) litres to \(3.00\) litres. Find the ratio of (a) the final pressure to the initial pressure and (b) the final temperature to the initial temperature.

A mixture contains 1 mole of helium \(\left(C_{p}=2 \cdot 5 R\right.\), \(C_{V}=1 \cdot 5 R\) ) and 1 mole of hydrogen \(\left(C_{p}=3 \cdot 5 R\right.\) \(C_{V}=2 \cdot 5 R\) ). Calculate the values of \(C_{p}, C_{V}\) and \(\gamma\) for the mixture.

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