Internal energy of a gas and kinetic energy, "typical velocity"

In summary, the internal energy of a gas is primarily determined by the kinetic energy of its molecules, which is influenced by temperature. The "typical velocity" of gas molecules refers to the average speed at which they move, which can be derived from the temperature and mass of the gas. As temperature increases, the kinetic energy and typical velocity of the gas molecules also increase, leading to higher internal energy.
  • #1
laser
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Homework Statement
See description
Relevant Equations
Kavg = 3/2kt
Source: Shankar Yale OCW physics
Screenshot_1.png

I have three questions here:

1. K_avg is 3/2kT, sure. But isn't this the kinetic energy of one particle only? So why isn't the answer multiplied by avogadro's number (because one mole).

2. When doing the "typical velocity" derivation, I noticed that they used the root mean squared formula to derive the expression there. But I would think "typical velocity" means the probable velocity, i.e. the velocity when dP/dv = 0 in the probability vs velocity curve.

3. Just wondering, as nitrogen is a diatomic gas, why doesn't the internal energy have a factor of 5/2 as opposed to 3/2?

Thanks!
 
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  • #2
laser said:
Homework Statement: See description
Relevant Equations: Kavg = 3/2kt

Source: Shankar Yale OCW physics
View attachment 345547
I have three questions here:

1. K_avg is 3/2kT, sure. But isn't this the kinetic energy of one particle only? So why isn't the answer multiplied by avogadro's number (because one mole).

2. When doing the "typical velocity" derivation, I noticed that they used the root mean squared formula to derive the expression there. But I would think "typical velocity" means the probable velocity, i.e. the velocity when dP/dv = 0 in the probability vs velocity curve.

3. Just wondering, as nitrogen is a diatomic gas, why doesn't the internal energy have a factor of 5/2 as opposed to 3/2?

Thanks!
  1. You are right and the answer is wrong. The average energy needs to be multiplied by Avogadro's number.
  2. "Typical" velocity is not typical use. It could mean r.m.s. velocity in the problem author's mind.
  3. Because the author of the solution apparently doesn't know the difference between the two or has calculated the internal energy of a diatomic molecule in the approximation ##2 \approx 1.##
Also note that the answer T = 244000 K is 10 times larger than the correct answer. The author of the solution has noted that it "is clearly a very high temperature" but has not considered that there might be a calculational error here.

I suggest that you reconsider this site as a source of learning.

Edited to fix problem with not considering the volume. See posts #5 and #6.
 
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  • #3
kuruman said:
Also note that the answer T = 244000 K is 10 times larger than the correct answer.
I think it is correct
 
  • #4
laser said:
I think it is correct
I don't think so.
##2\times 1.013\times 10^5=202,600##. If you divide by 8.314, you don't get ##244,000## which is a larger number. when multiplied by the volume gives the right answer.

Edited to fix problem with not considering the volume. See posts #5 and #6.
 
Last edited:
  • #5
kuruman said:
I don't think so.
##2\times 1.013\times 10^5=202,600##. If you divide by 8.314, you don't get ##244,000## which is a larger number.
volume is 10 m^3
 
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  • #6
Ah, yes. That's what I missed.
 
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FAQ: Internal energy of a gas and kinetic energy, "typical velocity"

What is internal energy in the context of a gas?

Internal energy refers to the total energy contained within a gas, which includes the kinetic energy of the gas molecules due to their motion, as well as potential energy associated with the interactions between the molecules. It is a state function that depends on the temperature, volume, and amount of substance in the gas.

How is kinetic energy related to the internal energy of a gas?

The kinetic energy of gas molecules is a significant component of the internal energy of the gas. The average kinetic energy of the molecules is directly proportional to the temperature of the gas. According to the kinetic theory of gases, the internal energy can be expressed as a function of the temperature, where higher temperatures correspond to higher average kinetic energies of the molecules.

What is "typical velocity" in the context of gas molecules?

Typical velocity, often referred to as the root-mean-square (RMS) velocity, is a measure of the average speed of gas molecules in a sample. It is calculated using the kinetic theory of gases and is influenced by the temperature and molar mass of the gas. The RMS velocity provides insight into the energy and motion of gas molecules.

How can the typical velocity of gas molecules be calculated?

The typical velocity (RMS velocity) of gas molecules can be calculated using the formula: \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the absolute temperature in Kelvin, and \( m \) is the mass of a gas molecule. Alternatively, it can also be expressed in terms of molar mass: \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the universal gas constant and \( M \) is the molar mass of the gas.

How does temperature affect the internal energy and typical velocity of a gas?

Temperature has a direct impact on both the internal energy and the typical velocity of a gas. As the temperature increases, the average kinetic energy of the gas molecules also increases, leading to higher internal energy. Consequently, the typical velocity of the gas molecules increases as well, resulting in faster-moving molecules at higher temperatures.

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