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Lecture Comments (10)

1 answer

Last reply by: Professor Dan Fullerton
Wed Aug 12, 2015 5:18 PM

Post by Anh Dang on August 12, 2015

Just to be sure, the gas constant R in the ideal gas equation is always going to be 8.31 J/(mol K)?

1 answer

Last reply by: Professor Dan Fullerton
Sat Apr 4, 2015 10:44 AM

Post by Jae Chang Lee on April 4, 2015

How do i find the molar mass of nitrogen and hydrogen in example 9?

1 answer

Last reply by: Professor Dan Fullerton
Fri Aug 1, 2014 10:31 AM

Post by Jamal Tischler on August 1, 2014

Isn't the internal energy i/2*n*R*T ? i is 3 for monoatomic, 5 for diatomic, 6 for poliatomic.

1 answer

Last reply by: Professor Dan Fullerton
Wed Jul 17, 2013 10:04 AM

Post by Andrew Buckelew on July 16, 2013

Isn't the Volume of ideal gas at STP 22.4 L and the V of a ideal gas at room temp. 24.4 L?

Love the videos! You do an amazing job!

0 answers

Post by Professor Dan Fullerton on July 9, 2013

It's a fundamental constant.  Avogadro's Number multiplied by Boltzmann's Constant gives you the universal gas constant (that's where R comes from).

0 answers

Post by KyungYeop Kim on July 9, 2013

Question: At 11:40, you say that U= 3/2 n*No*T implies that No*Kb=R. (That The Avogadro's number X Boltsman's constant = R, as in the ideal gas law.) But I don't understand. How does the former equation imply the latter? in other words, why do we assume that NoKb=R ? Is it a given as something to be memorized, or something that can be explained? (having a chemistry background I only know the R as in pv=nRT) Thank you in advance.

Ideal Gases

  • The temperature of a system characterizes the average kinetic energy of its molecules.
  • In an ideal gas, PV=nRT
  • Ideal gases have many particles moving randomly. The particles are far apart and do not exert forces upon each other unless they collide elastically.
  • The internal energy of an ideal gas is given by 3nRT/2.
  • The root mean square velocity is similar to an average velocity for molecules or atoms in a system.
  • The velocity of the particles in a gas varies widely and is characterized by a statistical distribution.

Ideal Gases

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:10
  • Ideal Gases 0:25
    • Gas Is Comprised of Many Particles Moving Randomly in a Container
    • Particles Are Far Apart From One Another
    • Particles Do Not Exert Forces Upon One Another Unless They Come In Contact in an Elastic Collision
  • Ideal Gas Law 1:18
  • Atoms, Molecules, and Moles 2:56
    • Protons
    • Neutrons
    • Electrons
    • Examples
  • Example 1: Counting Moles 4:58
  • Example 2: Moles of CO2 in a Bottle 6:00
  • Example 3: Pressurized CO2 6:54
  • Example 4: Helium Balloon 8:53
  • Internal Energy of an Ideal Gas 10:17
    • The Average Kinetic Energy of the Particles of an Ideal Gas
    • Total Internal Energy of the Ideal Gas Can Be Found by Multiplying the Average Kinetic Energy of the Gas's Particles by the Numbers of Particles in the Gas
  • Example 5: Internal Energy of Oxygen 12:00
  • Example 6: Temperature of Argon 12:41
  • Root-Mean-Square Velocity 13:40
    • This is the Square Root of the Average Velocity Squared For All the Molecules in the System
    • Derived from the Maxwell-Boltzmann Distribution Function
  • Calculating vrms 14:56
  • Example 7: Average Velocity of a Gas 18:32
  • Example 8: Average Velocity of a Gas 19:44
  • Example 9: vrms of Molecules in Equilibrium 20:59
  • Example 10: Moles to Molecules 22:25
  • Example 11: Relating Temperature and Internal Energy 23:22

Transcription: Ideal Gases

Hi everyone. I am Dan Fullerton and I would like to welcome you back to, as we continue our study of thermophysics and thermodynamics, by talking about ideal gases. 0000

Now our objectives for this lesson are going to be utilizing the ideal gas law to solve for pressure, volume, temperature and quality of an ideal gas, and explaining the relationship between root mean square velocity and the temperature of a gas. 0010

With that, let us talk about ideal gases. 0023

Ideal gases are theoretical models of real gases which utilize a number of basic assumptions to simplify their study. 0027

The first assumption is that the gas is comprised of many particles moving randomly in a container. 0034

One or two molecules in a container is not a really good model. 0039

We need to have some substantial amount of gas. 0042

The particles are on average, far apart from one another. 0045

They are not combined and almost in a liquid state. 0049

In the particles, do not exert forces upon one another unless they come in contact in an elastic collision. 0052

So we can neglect things like the gravitational force of attraction between these tiny particles. 0058

Now, this works well for most gases at standard temperatures and pressures, but it does not hold up so well for very heavy gases at low temperatures or very high pressures, but for most of the things we would want to use it for, it works just great. 0064

The ideal gas law relates pressure, volume, number of particles and temperature of an ideal gas in a single equation. 0079

You can see this written in a number of different forms. 0086

Pressure times volume equals (NRT) -- PV = NRT or NKbt depending on how you want to see it written and we will talk about what these values are. 0089

Now (n), the number of moles of a gas, (n), is (N), which is the number of molecules divided by Avogadro's number or 6.02 × 1023 something's per mole. 0100

In this equation, pressure (P) is given in Pascal's, the volume (V) is in m3, and if you use (n), that is the number of moles of a gas and we use that over here. 0114

(R), then is the universal gas constant or 8.31 J/mol K. 0126

Use that if we are using the number of moles in a gas version and (T) is the temperature in Kelvins. 0132

On the other hand, if you want to use this version, PV = NKbT, where (N) is the number of molecules that you have and Kb is Boltzmann's constant -- we talked about that previously -- 1.38 × 10-23 J/K, and T, again is your temperature in Kelvins. 0137

Now before we move on, it is probably important to note that one mole of a gas at standard temperature and pressure has a volume of just about 24 L. 0157

That is always a good thing just to have in the back of your mind. 0172

Atoms, molecules, and moles -- Atoms are made up of protons and we will call the number of protons the Z-number, so when we write something like a molecule or an atom, (x) is the symbol for it and (z) is the number of protons that goes down here to the bottom and to the left of the element. 0177

(N) the number of neutrons, they do not have a charge and if neutral, you have Z-electrons; you have one electron with a charge of -1 elementary charge for every proton. 0195

Now the atomic mass (A) is the number of protons plus neutrons, so you have the protons here and you have the protons plus neutrons here, so if you want to adjust the number of neutrons, take (A) subtract (Z) and you will be left with the number of neutrons you have for your atom molecule element. 0205

Here we have 2, 4 helium, that means that we have two protons; we have four protons and neutrons, which means we must have two neutrons and one mole of helium is approximately 4 grams (g). 0225

Here we have Carbon-14 -- the 6 tells you that it has six protons, the 14 means it has 6 protons and neutrons, therefore we must have 8 neutrons and one mole of this material, which has a mass of about 14 g. 0246

And if we looked at something like oxygen, that has 8 protons, 8 neutrons, and if we looked at one mole of molecular oxygen (O2), it is going to have a mass of about 32 g -- because it is O2, we have two of them there. 0268

And that is going to have 6.02 × 10 23 molecules. 0288

All right, so let us see how we can put some of this together. 0298

How many moles of an ideal gas are equivalent to 3.01 × 1024 molecules? 0301

Well, let us start with 3.01 × 1024 molecules and I am going to multiply that by one mole over 6.02 × 10 23 molecules in a mole, Avogadro's number. 0307

And really what I am doing is multiplying by 1 and anything I multiply by 1, I get the same value even if the units are changing. 0328

One mole and 6.02 × 10 23 molecules are really the same thing, so 1/1 = 1, however, when I do this multiplication, my molecules will cancel out and I will be left with units of moles. 0334

3.01 × 1024 × 1 divided by 6.02 × 10 23 = 5 moles. 0348

Let us look at another example -- For moles of carbon dioxide in a bottle, how many moles of gas are present in a 0.3 m3 bottle of carbon dioxide held at a temperature of 320 K and a pressure of 1 million Pa? 0360

We will use our ideal gas law, PV = NRT, therefore (N) the number of moles, is going to be equal to PV/RT, where our (P) is 1 million or 106 Pa, and (V) is 0.3 m3. 0376

(R), our universal gas constant is 8.31 and our (T) is 320 K. 0394

That gives me about 113 moles. 0402

Let us take a look at another example -- Pressurized carbon dioxide. 0414

We have a cubic meter of carbon dioxide gas at room temperature, 300 K, an atmospheric pressure at about 101,325 Pa, and it is compressed into a volume of 0.1 m3 and held at a temperature of 260 K. 0418

What is the pressure of this compressed carbon dioxide? 0431

Since the number of moles of gas is a constant here, we can simplify the ideal gas equation into some combined gas law by setting the initial pressure volume and temperature relationship equal to the final pressure volume in temperature relationship. 0435

If PV = NRT, and we are holding (N) and (R) constant, I could pull (T) over to this side for PV/T = NR, so NR must be constant. 0450

So I could write this as P1(V1)/T1 = P2(V2)/T2. 0466

And since I want P2, I can rearrange that and say P2 = P1(V1)T2/T1(V2). 0476

Now I can substitute in my values -- P2 = 101,325 Pa (P1) 13 (V1) 260 K (T2)/300 K (T1) 0.1 M3 (V2). 0493

If I do this, I come out with a P2 or final pressure of about 878,000 Pa. 0517

Let us look at a helium balloon. 0533

One mole of helium gas is placed inside a balloon. 0535

What is the pressure -- looking for pressure inside the balloon -- when the balloon rises to a point in the atmosphere where the temperature is -12 ° C and the volume of the balloon is 0.25 m4?0538

First thing is to convert this temperature from Celsius to Kelvins. 0551

Temperature in K is our temperature in ° C + 273.15, so that is going to be -12 ° C + 273.15 or 261.15 K. 0556

Now, if PV = NRT, then P = NRT/V. 0578

Well, (N), 1 mole; (R), the gas constant (8.31); our temperature (261.15 K)... 0589

...and our volume here (0.25 m3) gives us a pressure of about 8,680 Pa. 0600

As we talk about the internal energy of an ideal gas, we call it the average kinetic energy of the particles is described by the equation, average kinetic energy is 3/2 times Boltzmann's constant times the temperature in Kelvins. 0618

Now the total internal energy of the ideal gas can then be found by multiplying the average kinetic energy of the gases particles by the number of particles. 0632

The total internal energy (U) is going to be the number of particles, (N), times the average kinetic energy, but we can do a little bit of manipulation here. 0641

The total number of atoms, particles, (N), is going to be equal to the number of moles times Avogadro's number and the average kinetic energy is 3/2 KVT. 0653

So when I substitute those into my equation, the total internal energy is going to be equal to 3/2 × number of moles (Avogadro's number) × Boltzmann's constant × temperature. 0671

This implies then, however, Avogadro's number × Boltzmann's constant is our universal gas constant (R), so I am going to take that and replace it with (R) to write that the total internal energy (U) is 3/2(N) -- now I can place my (R) in there -- NRT. 0689

I have a formula for the total internal energy of an ideal gas. 0712

Let us see how we can use that. 0719

Find the internal energy of 5 moles of oxygen at a temperature of 300 K. 0722

U = 3/2 (NRT), so that's 3/2 × 5 moles × our universal gas constant, 8.31 × 300 K or about 18,700 J or 18.7 kJ.0729

Let us do another one.0759

What does the temperature of 20 moles of argon with the total internal energy of 100 kJ?0761

Well, total internal energy (U) is 3/2 (NRT), therefore temperature equals 2 × the total internal energy divided by 3 × the number of moles × that universal gas constant (R).0768

So that's 2 × 100 kJ or 100,000 J divided by 3 × 20 moles × our universal gas constant, 8.31...0786

...which gives me about 401 K.0799

Great. Let us look a little bit more at the velocity of these particles.0817

The root means square velocity or (RMS) velocity is the square root of the average velocity squared for all the molecules in the system.0822

You can kind of think of it as a sort of average velocity for molecules when we are using this Maxwell Boltzmann distribution statistics, probabilistic statistics.0831

What we have down here, is we have a plot of number of atoms or molecules -- number of particles with some specific velocity for different materials at about 293.15 K -- closing in on room temperature. 0842

By the way, that C4H10, that is butane.0860

You can see that we have different spreads here.0863

For butane, we have a peak here at something just shy of 300 m/s0867

That is where you are going to have the most particles, but you have a fairly tight distribution around that.0876

As we go to something like ammonia, NH3, we have a much wider distribution and a greater tail down here at the higher velocities.0881

So, just an idea, giving you a feel for what root means square velocity is and what it means.0889

Calculating the root means square velocity -- We are going to start with the average kinetic energy as 3 1/2 × Boltzmann's constant × the temperature in Kelvins.0896

What this means then, is average kinetic energy -- is we are taking the average of 1/2MV2 for all those particles and that is equal to 3/2 × Boltzmann's constant × the temperature (T).0905

But the mass of these particles is constant, so taking the average of it we can pull the (M) out of the average and multiply it and we are done and so can the 1/2, that is a constant too.0921

That implies if we pull the M/2 out that M/2 × the average of V2 = 3/2 Kbt.0931

Or if I divide both sides by M/2 or think of it as multiplying both sides by 2/M, the left hand side is just going to be the average of V2 and the right hand side we are going to have 3/M Boltzmann's Constant × (T).0947

If I take the square root of both sides -- well, this is the definition of the average of the VRMS -- the root means square velocity.0970

So VRMS = the square root of 3/MKb × (T).0981

But this (M), the mass, we can give another symbol that is often used. 0992

(M) is often written as the mass of the molecule -- is written as μ.0998

So I could write VRMS = 3Kbt/μ square root.1003

There is another equation for the root means square velocity, but we can take that even further.1014

If we start with the root means square velocity equal to the square root of 3 Kbt/μ...1022

...this implies then, knowing that Boltzmann's Constant, Kb is actually R/Na (Avogadro's number), that we could write VRMS, our root means square velocity as equal to the square root of 3. 1032

Now we have our R/Na right there and we still our μ down here and we still have our (T). 1055

So we have 3RT/μ × Na, but even more, μ × Na, the mass of our molecules × Avogadro's number is going to give us what is known as the molar mass, (M) in kilograms per mole. 1067

So a little bit more we can do here. We could write this then as VRMS = the square root of 3RT/M -- another version for calculating the root means square velocity.1087

Let us take a look at an example here.1113

An ideal gas is placed in a closed bottle and cooled to half its original temperature. 1114

What happens to the average speed of the molecules?1119

Well, the root means square velocity is the square root of 3RT/M and we are going to cut (T) in half.1123

Everything else is going to stay the same, but if we cut (T) in half and (VRMS) is proportional to -- well that is going to be -- if (T) is 1/2 of what it was, that is going be proportional to square root of 1/2 -- what we had for its original velocity.1139

Square root of 1/2 is about 0.71, so it is going to be 0.71 of its original velocity or you could write this as the average speed -- if we think of it in terms of average speed -- is going to be about 71% its original value.1158

Take its original value multiplied by 0.71.1177

Let us do another one. These can be a little bit tricky when you see them the first time.1180

The root means square velocity of the molecules of a 300 K gas is 1000 m/s. 1184

What is the root means square velocity of the molecules at 600 K?1194

Well, again we will start with VRMS = square root of 3RT/M.1199

Now we are going to double the temperature and when we double the temperature its proportional to the square root of (T), so (VRMS) is proportional to the square root of 2 times the original (RMS) velocity.1210

So that is going to be square root of 2 × 1,000 m/s or VRMS = 1.41, the original, which is 1410 m/s. 1228

You get a 41% increase and the root means square velocity of the molecules and you double the temperature.1245

All right, trying another one -- Hydrogen (H2) and Nitrogen (N2) gas are in thermal equilibrium in a closed box. 1259

Compare the root means square velocities of the molecules.1267

Well, we are going to start by referencing our (VRMS) equation is equal to the square root of 3RT/M.1271

Now the (M) of hydrogen is 2 and the (M) of nitrogen is 28. 1282

That means we have a 14 times difference.1291

All right, so when I look at what these are proportional to it is 1/M, so if I were to take a ratio of these two, at the top I would the square root of 1/2 because we have 2 for the (M) of hydrogen compared to 1/square root of 28 for my ratio for nitrogen.1297

Which is going give me an (x) factor of 3.74.1320

That means the root means square velocity for hydrogen is going to be 3.74 times larger than the root means square velocity for nitrogen.1326

That has to be expected; it is a lot smaller.1339

Find the number of molecules in 0.4 moles of an ideal gas. 1346

All right, a conversion problem -- 0.4 moles -- and we want to convert this into molecules.1350

I am going to multiply and I want moles to go away, so I will put that in the denominator so they make a ratio of 1 and cancel out.1360

I want molecules as my unit and now I need to make sure I am multiplying by a value of 1.1366

One mole is equal to 6.02 × (10)23 molecules. 1372

My moles, units, will make your ratio of 1 or cancel out and I will be left with 0.4 × 6.02 × 1023 molecules, which is 2.4 × (10)23 molecules.1378

All right, let us look at one last problem here.1401

The temperature of an ideal gas is doubled. 1405

What happens to its internal energy?1408

The first thing I am going to do is recall that internal energy equation, U 3/2 NRT.1411

Now if I double the temperature -- All right if I am doubling the temperature here, I must be doubling the internal energy and I get double the internal energy.1420

So the short answer -- internal energy doubles.1433

Hopefully that gets you a good start on ideal gases.1445

Thank you so much for your time coming to

Make it a great day everyone!1454