B Energy of Ideal Gas: Internal E & Kinetic E

AI Thread Summary
The discussion centers on the calculation of internal energy for an ideal gas, specifically questioning the exclusion of electrostatic energy in the kinetic energy formula K.E = 3/2 n R T. It emphasizes that this formula applies only to monatomic gases and does not account for potential energy or interactions in diatomic or polyatomic gases. The analogy to snooker balls illustrates that, in the ideal gas model, particles are treated as non-interacting entities. The limitations of the model are acknowledged, particularly under conditions of high pressure and low temperature. Overall, while the ideal gas approximation is useful, it has significant constraints that must be recognized.
Bhope69199
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Hi,

When calculating the energy of an ideal gas we neglect the potential energy and calculate the kinetic energy using:

K.E = 3 /2 n R T

My question is why do we not consider the electrostatic energy of the gas?

If I am trying to work out the internal energy of 1 mol of Radon, why do I only require the kinetic energy and not the energy that was required to bring the electrons and protons together? Is this energy not held within the gas?
 
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That's the meaning of ideal gas, a gas whose particles don't interact with each other. Its an approximation! Of course you can consider interactions and have a more precise model, but that means you should do harder calculations.
 
If your gas is so hot that electrons can be removed from atoms (=your gas is actual a plasma), you cannot approximate it as ideal gas any more.
 
Bhope69199 said:
Hi,

When calculating the energy of an ideal gas we neglect the potential energy and calculate the kinetic energy using:

K.E = 3 /2 n R T

My question is why do we not consider the electrostatic energy of the gas?

If I am trying to work out the internal energy of 1 mol of Radon, why do I only require the kinetic energy and not the energy that was required to bring the electrons and protons together? Is this energy not held within the gas?
When snooker balls collide it is not necessary to consider the "electrostatic energy" of the snooker balls..molecules are considered to be 'snooker balls' in the simple kinetic theory of gases
 
Bhope69199 said:
Hi,

When calculating the energy of an ideal gas we neglect the potential energy and calculate the kinetic energy using:

K.E = 3 /2 n R T
Be aware that this formula is only valid for a monatomic ideal gas. It is not valid if the gas molecules have 2 or more atoms -- e.g. O2, N2, CO2, etc.

I've personally known physics teachers who were unaware of this.
 
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Redbelly98 said:
Be aware that this formula is only valid for a monatomic ideal gas. It is not valid if the gas molecules have 2 or more atoms -- e.g. O2, N2, CO2, etc.

I've personally known physics teachers who were unaware of this.

oops...should perhaps have qualified 'translational kinetic energy'
this is more or less implied among physicists when the model is based on snooker balls, as far as I know there are no diatomic snooker balls.
Such a simple model has many limitations but is surprisingly informative.
The model also breaks down at high pressures and low temperatures, I imagine most physics teachers would be aware of this or at least suspect it.
 
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