Temperature of a gas in a container

In summary, the conversation discusses a rectangular container with gas inside that accelerates in free space. Despite the increase in kinetic energy, the temperature of the gas does not change. There are two possible explanations for this: 1) the impulsive force imparted by the moving container on the gas molecules causes them to move in a disciplined manner that does not significantly change the Maxwellian distribution, or 2) with uniform motion, there are no changes in kinetic energy.
  • #1
physicsarya
1
0
A rectangular (simplified) container with rigid surfaces, has a certain mass of ideal gas within it, and it accelerates in free space, undergoing rectilinear motion. There are no dissipative forces. Now, since the container moves, its kinetic energy increases, and since the temperature of the enclosed gas is dependent on the kinetic energy, their temperature should consequently increase, which does not occur. What's the reason behind it?

I try to solve it in the following manner:

Lets consider a rectangular container and let's label its vertical edges A and B. Assuming that it moves along BA ,after a certain time the edge A moves to A', B to B', and thus, the effective space for the molecules remains always unchanged. Now considering the force parameter in the problem, here the forcing agent is only the edge B colliding with the molecules in the vicinity(let vicinity be the region BC, C being sufficiently close to B) of B. Now, I argue that the impulsive force imparted by B on the molecules near it, modifies the initial random motion direction of these nearby molecules, and forces them to move along BA direction. But after BC, i.e. in the region CA, the molecules don't attain a perfectly horizontal motion, and the alignment along BA deteriorates as one moves from B to A, in a disciplined manner, and yet not changing the Maxwellian distribution. I consider layers within the container, at the not-so-clearly defined C junction, there's a continuous yet slightly increasing variation from BA direction, although the resultant velocity of these molecules may be in NW/SW directions. These molecules collide with those in the next layer, and these latter molecules undergo further variation from BA (not to forget that in all these layers, some molecules might even move in NE/SE/E(AB) directions, the fraction of which is relatively low, but increases, and becomes relatively higher while moving along BA). Thus this variation considerably increases in the vicinity of the A edge, and as such, magnitudes of the velocities of these molecules eventually re-organize themselves (due to the changing angle-factor at each layer of molecules) in such a way that the final energy-distribution curve varies too little from the initial one, to be of any significance at a practical level, whence the temperature of the enclosed gas remains unchanged.

Is this a correct approach, as far as an explanation is concerned?

P.S. also provide me with an alternative simpler explanation, which I am sure there is...
 
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  • #2
physicsarya said:
since the container moves, its kinetic energy increases, and since the temperature of the enclosed gas is dependent on the kinetic energy, their temperature should consequently increase, which does not occur

Be carefull and specify clearly what is the system that you want to analyse: container plus gas inside or only gas inside. With uniform motion there are no changes of kinetic energy.
 

1. What factors affect the temperature of a gas in a container?

The temperature of a gas in a container is affected by several factors, including the volume of the container, the number of gas particles present, and the amount of energy (or heat) added or removed from the system. The type of gas and its properties, such as its specific heat capacity, also play a role in determining the temperature.

2. How does the ideal gas law relate to the temperature of a gas in a container?

The ideal gas law, which states that pressure, volume, and temperature are directly proportional for a given amount of gas, can be used to calculate the temperature of a gas in a container. By manipulating the equation (PV = nRT), we can solve for temperature (T) by dividing the pressure (P) by the product of the number of moles of gas (n) and the gas constant (R).

3. Why does the temperature of a gas increase when it is compressed in a container?

When a gas is compressed in a container, its volume decreases, causing the gas particles to collide more frequently with each other and the walls of the container. This increases the kinetic energy of the particles, resulting in a higher temperature. Additionally, the compression of the gas can also increase the pressure, which can further contribute to an increase in temperature.

4. How does the temperature of a gas in a container change when heat is added or removed?

When heat is added to a gas in a container, the gas particles absorb this energy and their kinetic energy increases, thereby increasing the temperature. Conversely, when heat is removed from a gas in a container, the gas particles lose energy and their kinetic energy decreases, resulting in a decrease in temperature. This relationship is described by the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred and transformed.

5. Can the temperature of a gas in a container be lower than absolute zero?

No, the temperature of a gas in a container cannot be lower than absolute zero, which is the theoretical lowest temperature possible (0 Kelvin or -273.15 degrees Celsius). At this temperature, the gas particles have minimal kinetic energy and cannot decrease in temperature any further. This concept is based on the third law of thermodynamics, which states that the entropy (or disorder) of a perfect crystal at absolute zero is zero.

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