WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone. I am Dan Fullerton and I would like to welcome you back to Educator.com, as we continue our study of thermophysics and thermodynamics, by talking about ideal gases.
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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.
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With that, let us talk about ideal gases.
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Ideal gases are theoretical models of real gases which utilize a number of basic assumptions to simplify their study.
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The first assumption is that the gas is comprised of many particles moving randomly in a container.
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One or two molecules in a container is not a really good model.
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We need to have some substantial amount of gas.
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The particles are on average, far apart from one another.
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They are not combined and almost in a liquid state.
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In the particles, do not exert forces upon one another unless they come in contact in an elastic collision.
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So we can neglect things like the gravitational force of attraction between these tiny particles.
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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.
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The ideal gas law relates pressure, volume, number of particles and temperature of an ideal gas in a single equation.
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You can see this written in a number of different forms.
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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.
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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 × 10^23 something's per mole.
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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.
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(R), then is the universal gas constant or 8.31 J/mol K.
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Use that if we are using the number of moles in a gas version and (T) is the temperature in Kelvins.
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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.
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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.
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That is always a good thing just to have in the back of your mind.
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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.
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(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.
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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.
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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).
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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.
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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.
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And that is going to have 6.02 × 10 ^23 molecules.
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All right, so let us see how we can put some of this together.
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How many moles of an ideal gas are equivalent to 3.01 × 10^24 molecules?
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Well, let us start with 3.01 × 10^24 molecules and I am going to multiply that by one mole over 6.02 × 10 ^23 molecules in a mole, Avogadro's number.
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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.
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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.
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3.01 × 10^24 × 1 divided by 6.02 × 10 ^23 = 5 moles.
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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?
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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 10^6 Pa, and (V) is 0.3 m³.
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(R), our universal gas constant is 8.31 and our (T) is 320 K.
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That gives me about 113 moles.
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Let us take a look at another example -- Pressurized carbon dioxide.
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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.
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What is the pressure of this compressed carbon dioxide?
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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.
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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.
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So I could write this as P1(V1)/T1 = P2(V2)/T2.
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And since I want P2, I can rearrange that and say P2 = P1(V1)T2/T1(V2).
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Now I can substitute in my values -- P2 = 101,325 Pa (P1) 1³ (V1) 260 K (T2)/300 K (T1) 0.1 M³ (V2).
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If I do this, I come out with a P2 or final pressure of about 878,000 Pa.
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Let us look at a helium balloon.
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One mole of helium gas is placed inside a balloon.
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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 m^4?
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First thing is to convert this temperature from Celsius to Kelvins.
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Temperature in K is our temperature in ° C + 273.15, so that is going to be -12 ° C + 273.15 or 261.15 K.
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Now, if PV = NRT, then P = NRT/V.
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Well, (N), 1 mole; (R), the gas constant (8.31); our temperature (261.15 K)...
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...and our volume here (0.25 m³) gives us a pressure of about 8,680 Pa.
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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.
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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.
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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.
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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.
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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.
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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.
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I have a formula for the total internal energy of an ideal gas.
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Let us see how we can use that.
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Find the internal energy of 5 moles of oxygen at a temperature of 300 K.
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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.
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Let us do another one.
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What does the temperature of 20 moles of argon with the total internal energy of 100 kJ?
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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).
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So that's 2 × 100 kJ or 100,000 J divided by 3 × 20 moles × our universal gas constant, 8.31...
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...which gives me about 401 K.
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Great. Let us look a little bit more at the velocity of these particles.
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The root means square velocity or (RMS) velocity is the square root of the average velocity squared for all the molecules in the system.
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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.
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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.
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By the way, that C4H^10, that is butane.
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You can see that we have different spreads here.
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For butane, we have a peak here at something just shy of 300 m/s
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That is where you are going to have the most particles, but you have a fairly tight distribution around that.
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As we go to something like ammonia, NH3, we have a much wider distribution and a greater tail down here at the higher velocities.
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So, just an idea, giving you a feel for what root means square velocity is and what it means.
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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.
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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).
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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.
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That implies if we pull the M/2 out that M/2 × the average of V² = 3/2 Kbt.
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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).
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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.
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So VRMS = the square root of 3/MKb × (T).
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But this (M), the mass, we can give another symbol that is often used.
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(M) is often written as the mass of the molecule -- is written as μ.
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So I could write VRMS = 3Kbt/μ square root.
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There is another equation for the root means square velocity, but we can take that even further.
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If we start with the root means square velocity equal to the square root of 3 Kbt/μ...
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...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.
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Now we have our R/Na right there and we still our μ down here and we still have our (T).
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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.
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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.
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Let us take a look at an example here.
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An ideal gas is placed in a closed bottle and cooled to half its original temperature.
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What happens to the average speed of the molecules?
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Well, the root means square velocity is the square root of 3RT/M and we are going to cut (T) in half.
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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.
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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.
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Take its original value multiplied by 0.71.
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Let us do another one. These can be a little bit tricky when you see them the first time.
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The root means square velocity of the molecules of a 300 K gas is 1000 m/s.
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What is the root means square velocity of the molecules at 600 K?
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Well, again we will start with VRMS = square root of 3RT/M.
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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.
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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.
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You get a 41% increase and the root means square velocity of the molecules and you double the temperature.
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All right, trying another one -- Hydrogen (H2) and Nitrogen (N2) gas are in thermal equilibrium in a closed box.
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Compare the root means square velocities of the molecules.
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Well, we are going to start by referencing our (VRMS) equation is equal to the square root of 3RT/M.
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Now the (M) of hydrogen is 2 and the (M) of nitrogen is 28.
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That means we have a 14 times difference.
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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.
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Which is going give me an (x) factor of 3.74.
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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.
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That has to be expected; it is a lot smaller.
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Find the number of molecules in 0.4 moles of an ideal gas.
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All right, a conversion problem -- 0.4 moles -- and we want to convert this into molecules.
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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.
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I want molecules as my unit and now I need to make sure I am multiplying by a value of 1.
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One mole is equal to 6.02 × (10)^23 molecules.
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My moles, units, will make your ratio of 1 or cancel out and I will be left with 0.4 × 6.02 × 10^23 molecules, which is 2.4 × (10)^23 molecules.
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All right, let us look at one last problem here.
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The temperature of an ideal gas is doubled.
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What happens to its internal energy?
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The first thing I am going to do is recall that internal energy equation, U 3/2 NRT.
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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.
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So the short answer -- internal energy doubles.
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Hopefully that gets you a good start on ideal gases.
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Thank you so much for your time coming to Educator.com.
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Make it a great day everyone!