Can someone help me make the connection between KEavg=3/2kT, to KEtotal=3/2nRT

Are they separate formulas taught in physics or do we derive the KEtotal through other formulas?

I understand Boltsman(k)=R/Na

And KE=1/2MV² is supposed to help but I’m not sure how.

Thank you

Are they separate formulas taught in physics or do we derive the KEtotal through other formulas?

I understand Boltsman(k)=R/Na

And KE=1/2MV² is supposed to help but I’m not sure how.

Thank you

Original post by KingRich

Can someone help me make the connection between KEavg=3/2kT, to KEtotal=3/2nRT

Are they separate formulas taught in physics or do we derive the KEtotal through other formulas?

I understand Boltsman(k)=R/Na

And KE=1/2MV² is supposed to help but I’m not sure how.

Thank you

Are they separate formulas taught in physics or do we derive the KEtotal through other formulas?

I understand Boltsman(k)=R/Na

And KE=1/2MV² is supposed to help but I’m not sure how.

Thank you

3/2 nRT is the energy of n moles of gas. Essentially, yes, you are using k = R/Na, but you also are multiplying it by the number of moles there are.

Original post by UtterlyUseless69

3/2 kT is the energy of a single particle.

3/2 nRT is the energy of n moles of gas. Essentially, yes, you are using k = R/Na, but you also are multiplying it by the number of moles there are.

3/2 nRT is the energy of n moles of gas. Essentially, yes, you are using k = R/Na, but you also are multiplying it by the number of moles there are.

So, is the formula 3/2nRT a formula taught in physics?

Or, is it something that has to be derived and shown?

(edited 1 month ago)

Original post by KingRich

So, is the formula 3/2nRT a formula taught in physics?

Or, is it something that has to be derived and shown?

Or, is it something that has to be derived and shown?

I suppose you can derive or show it starting from 3/2 kT, but I doubt they’d really care if you just quoted it without proof.

Also, I realise I forgot to address your most important question.

1/2 mv^2 is related to these formulae through something called “equipartition theorem”. Unless you are doing degree level chemistry/physics, I doubt you need to understand it, but here goes:

Equipartition theorem boils down to something along the lines of “for every quadratic degree of freedom a gas particle has to move, there is a 1/2 kT (or 1/2 RT for bulk matter) contribution to the (translational) kinetic energy”.

Gases move in 3 dimensions - call these x, y and z. Suppose the components of the velocity of the gas are v(x), v(y) and v(z).

The total kinetic energy will be 1/2 mv(x)^2 + 1/2 mv(y)^2 + 1/2 mv(z)^2 = 1/2 m(v(x)^2 + v(y)^2 + v(z)^2) - this is three quadratic terms, so the kinetic energy should be 3 x 1/2 kT = 3/2 kT for a single particle.

In truth, this only holds for monatomic gases as gases with more atoms can have rotational (and vibrational, if the temperature is high enough) contributions to the number of quadratic degrees of freedom.

1/2 mv^2 is related to these formulae through something called “equipartition theorem”. Unless you are doing degree level chemistry/physics, I doubt you need to understand it, but here goes:

Equipartition theorem boils down to something along the lines of “for every quadratic degree of freedom a gas particle has to move, there is a 1/2 kT (or 1/2 RT for bulk matter) contribution to the (translational) kinetic energy”.

Gases move in 3 dimensions - call these x, y and z. Suppose the components of the velocity of the gas are v(x), v(y) and v(z).

The total kinetic energy will be 1/2 mv(x)^2 + 1/2 mv(y)^2 + 1/2 mv(z)^2 = 1/2 m(v(x)^2 + v(y)^2 + v(z)^2) - this is three quadratic terms, so the kinetic energy should be 3 x 1/2 kT = 3/2 kT for a single particle.

In truth, this only holds for monatomic gases as gases with more atoms can have rotational (and vibrational, if the temperature is high enough) contributions to the number of quadratic degrees of freedom.

(edited 1 month ago)

Original post by UtterlyUseless69

Also, I realise I forgot to address your most important question.

1/2 mv^2 is related to these formulae through something called “equipartition theorem”. Unless you are doing degree level chemistry/physics, I doubt you need to understand it, but here goes:

Equipartition theorem boils down to something along the lines of “for every quadratic degree of freedom a gas particle has to move, there is a 1/2 kT (or 1/2 RT for bulk matter) contribution to the (translational) kinetic energy”.

Gases move in 3 dimensions - call these x, y and z. Suppose the components of the velocity of the gas are v(x), v(y) and v(z).

The total kinetic energy will be 1/2 mv(x)^2 + 1/2 mv(y)^2 + 1/2 mv(z)^2 = 1/2 m(v(x)^2 + v(y)^2 + v(z)^2) - this is three quadratic terms, so the kinetic energy should be 3 x 1/2 kT = 3/2 kT for a single particle.

In truth, this only holds for monatomic gases as gases with more atoms can have rotational (and vibrational, if the temperature is high enough) contributions to the number of quadratic degrees of freedom.

1/2 mv^2 is related to these formulae through something called “equipartition theorem”. Unless you are doing degree level chemistry/physics, I doubt you need to understand it, but here goes:

Equipartition theorem boils down to something along the lines of “for every quadratic degree of freedom a gas particle has to move, there is a 1/2 kT (or 1/2 RT for bulk matter) contribution to the (translational) kinetic energy”.

Gases move in 3 dimensions - call these x, y and z. Suppose the components of the velocity of the gas are v(x), v(y) and v(z).

The total kinetic energy will be 1/2 mv(x)^2 + 1/2 mv(y)^2 + 1/2 mv(z)^2 = 1/2 m(v(x)^2 + v(y)^2 + v(z)^2) - this is three quadratic terms, so the kinetic energy should be 3 x 1/2 kT = 3/2 kT for a single particle.

In truth, this only holds for monatomic gases as gases with more atoms can have rotational (and vibrational, if the temperature is high enough) contributions to the number of quadratic degrees of freedom.

I see. The course I’m doing is an access engineering course. Their grades are between, distinction, merit and pass. So, my thinking is. The more I can show, the happier they’ll be.

You have been great help.

Out of curiosity, as I’ve mostly been learning A-level maths up until now, using units in the question were never really important but I saw that in physics itself, the units are included in the question? Is this still true today?

Original post by KingRich

I see. The course I’m doing is an access engineering course. Their grades are between, distinction, merit and pass. So, my thinking is. The more I can show, the happier they’ll be.

You have been great help.

Out of curiosity, as I’ve mostly been learning A-level maths up until now, using units in the question were never really important but I saw that in physics itself, the units are included in the question? Is this still true today?

You have been great help.

Out of curiosity, as I’ve mostly been learning A-level maths up until now, using units in the question were never really important but I saw that in physics itself, the units are included in the question? Is this still true today?

You may need to do dimensional analysis to check your units (i.e plugging just the relevant units into the formula and seeing if it works). In the case of energy, it should be in J (or kg m^2 s^-2 as a derived unit). As such, for the formula KE = 1/2 mv^2 to work, you really need the mass to be in kg and the velocity to be in m s^-1:

KE has units (kg m^2 s^-2)

1/2 has no units (i.e call the units 1)

m has units (kg)

v has units (m s^-1)

KE = 1/2 x m x v^2

kg m^2 s^-2 = 1 x kg x (m s^-1)^2

kg m^2 s^-2 = kg m^2 s^-2

The statement is true, so the units of m must be kg and the units of v must be m s^-1.

(edited 1 month ago)

Original post by UtterlyUseless69

The data given to you in a question always should include units for A level physics - if you tried plugging a mass in grams and a velocity in miles per hour into KE = 1/2 mv^2, you most certainly would not get the energy in any meaningful unit.

You may need to do dimensional analysis to check your units (i.e plugging just the relevant units into the formula and seeing if it works). In the case of energy, it should be in J (or kg m^2 s^-2 as a derived unit). As such, for the formula KE = 1/2 mv^2 to work, you really need the mass to be in kg and the velocity to be in m s^-1:

KE has units (kg m^2 s^-2)

1/2 has no units (i.e call the units 1)

m has units (kg)

v has units (m s^-1)

KE = 1/2 x m x v^2

kg m^2 s^-2 = 1 x kg x (m s^-1)^2

kg m^2 s^-2 = kg m^2 s^-2

The statement is true, so the units of m must be kg and the units of v must be m s^-1.

You may need to do dimensional analysis to check your units (i.e plugging just the relevant units into the formula and seeing if it works). In the case of energy, it should be in J (or kg m^2 s^-2 as a derived unit). As such, for the formula KE = 1/2 mv^2 to work, you really need the mass to be in kg and the velocity to be in m s^-1:

KE has units (kg m^2 s^-2)

1/2 has no units (i.e call the units 1)

m has units (kg)

v has units (m s^-1)

KE = 1/2 x m x v^2

kg m^2 s^-2 = 1 x kg x (m s^-1)^2

kg m^2 s^-2 = kg m^2 s^-2

The statement is true, so the units of m must be kg and the units of v must be m s^-1.

Sorry, I meant just in general for physics questions. In regards to the solutions.

For example, using Boyles law to find the volume if pressure of a gas is increased…

P=101.345mmHg and V=5L

So, P₁V₁=P₂V₂

Do you work with just numbers? Or, include the units?

This is just a made up question by the way.

(edited 1 month ago)

Original post by KingRich

Sorry, I meant just in general for physics questions. In regards to the solutions.

For example, using Boyles law to find the volume if pressure of a gas is increased…

P=101.345mmHg and V=5L

So, P₁V₁=P₂V₂

Do you work with just numbers? Or, include the units?

This is just a made up question by the way.

For example, using Boyles law to find the volume if pressure of a gas is increased…

P=101.345mmHg and V=5L

So, P₁V₁=P₂V₂

Do you work with just numbers? Or, include the units?

This is just a made up question by the way.

You can just plug numbers in without needing to put the units in - just know that in the example you have given, all units must match (e.g both volumes will be in L if you take V1 = 5 L and both pressures will be in mmHg if you take p1 = 101.345 mmHg) and so you will need to convert any other data you have to a matching unit.

(edited 1 month ago)

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