Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a gas sitting at rest in a closed box. Add up the velocities of all its molecules, and the average velocity comes out zero. So, are the molecules standing still? Today we work through the concept checks of this unit's problem bank, starting with this one. Checks like these come up often in JEE, in statement form. So, let's begin.

0:27Check 1: zero average velocity, nonzero rms speed?

So, check one, our puzzle from the start. In a gas at rest, the molecules move at random, so their velocities point in all directions. Okay? Now the check. Can the average velocity of the molecules be zero, while their r m s speed, the root mean square speed, is not zero? Velocities have direction, and here they point in all directions, so they cancel in the average. On screen, v with a little arrow on top is the velocity, a vector, and the bar over it means the average. So the average velocity is zero. Right? But now square each speed. v squared, with the small two up top, is greater than zero for every moving molecule; the pointed sign means greater than. So the average of v squared, with the bar over the whole v squared, is greater than zero too. With me? And v r m s, the v with a small subscript r m s, is the square root of that average. So it is not zero; the equals sign with a slash through it means not equal. So yes, it can: the molecules race around in all directions.

1:37Check 2: a container of gas on the move

Clear? Now check two. Take a closed container of gas, and move it at a constant velocity, so every molecule gets the extra velocity of the container. So does T, the temperature, go up? Oh no, careful here, the temperature does not change at all. Temperature measures the random motion of the molecules, relative to the centre of mass of the gas. Okay? Moving the container adds the same velocity to every molecule: a uniform bulk velocity. The average velocity is now the container's velocity, but the random motion on top of it is untouched. Right? The one exception: if the container is stopped suddenly, its ordered kinetic energy becomes internal energy, and then the gas does get hotter. Got it?

2:25Check 3: why is Cp greater than Cv?

Now check three. C v and C p, capital C with a small v or a small p below, are the heat needed per mole, per kelvin: C v at constant volume, capital V, and C p at constant pressure, small p, okay? Here is the check. Same gas, same one kelvin rise, so why should C p be greater than C v? Start with constant volume. The gas cannot expand, so it does no work, and all the heat goes into raising U, the internal energy. Right? Now constant pressure. As the gas warms, it expands, so it also does work: p delta V, the pressure times delta V, the change in volume; delta is the little triangle. With me? But the rise in U is the same in both, because for an ideal gas, U depends on temperature alone. So at constant pressure we need extra heat for the work, and C p is greater than C v. Clear? And the difference is exactly R, the gas constant: C p minus C v equals R, per mole. Okay?

3:34Check 4: why did the Earth lose its hydrogen?

Now check four. Why has the Earth lost most of its hydrogen, but kept its nitrogen? At the same temperature, speeds go as one over the root of capital M, the molar mass. So v for hydrogen, subscript H two, over v for nitrogen, subscript N two, is the root of capital M for nitrogen over capital M for hydrogen. That is the root of twenty eight over two, which is root fourteen, about three point seven; the wavy equals sign means approximately. With me? So hydrogen moves about root fourteen times faster, and a much larger fraction of it exceeds the escape speed, the speed needed to leave the Earth for good. Hydrogen escapes, and nitrogen stays. Got it?

4:20Example 1: mean free path of helium vs xenon at same p, T

Now, Example one, the easier one. At the same pressure and temperature, which has the longer mean free path, helium or xenon? Pause the video, and try it first. Done? Okay, let's check. In the figure, one molecule moves in straight lines, and each orange dot is a collision. Each straight piece is a free path, and their average length is the mean free path, lambda, the Greek letter at the top. Right? Here, lambda equals k T, over root two, times pi, times d squared, times p. Small k is Boltzmann's constant, d is the molecular diameter, and small p is the pressure. The figure writes the same thing with n, the number density. With me? Now, helium atoms are much lighter, so they move much faster, and they hit other atoms more often. So helium has the shorter free path, and xenon the longer one, correct? Oh no, careful here, look at the formula. At the same p and T, only d can change. So lambda is proportional to one over d squared; the sign like an open alpha means proportional to. Okay? There is no mass and no speed in lambda. Faster atoms have more collisions each second, yes, but the mean free path itself does not depend on their speed. Right? So the answer is helium. Its atoms are smaller, so d is smaller, and lambda is longer. That's it!

5:52Example 2: is rms speed equal to the speed of sound?

Now Example two, a little harder. Is the r m s speed of the molecules equal to the speed of sound in the gas? Here is a hint: write both formulas side by side. Pause the video, and try it first. Done? Okay, let's check. v r m s is the root of three R T over capital M, the molar mass. And v s, the v with a small s below, is the speed of sound: the root of gamma R T over capital M, where gamma, the Greek letter like a little y, is C p over C v. Okay? Same gas, same temperature, and both formulas have R T over capital M under the root. So the two speeds are equal, correct? No, wait, look at what sits in front of R T: three in v r m s, but gamma in v s. Divide one by the other, and R, T and capital M all cancel. So v r m s over v s equals the root of three over gamma. Right? So everything depends on gamma. Is gamma more than three, or less? It is less, much less than three. Gamma equals one plus two over f, where f is the number of degrees of freedom, and the smallest f is three, for a monatomic gas. The pointed signs with a line below mean greater than or equal to, and less than or equal to. So gamma is at most five thirds, which is less than three. With me? So three over gamma is more than one, and so is its root. That means v r m s is greater than v s: the molecules move faster than sound in the gas. So the answer is no, they are not equal. Got it?

7:40Quick revision: Cp - Cv = R, mean free path, rms vs sound

So, quick revision. Temperature sees only random motion: the average velocity can be zero while v r m s is not, and bulk motion leaves the temperature alone. C p minus C v equals R, because at constant pressure the gas also does work; and light hydrogen escapes. At fixed p and T, lambda goes as one over d squared, and v r m s over v s is the root of three over gamma, more than one. Now try the practice question on this page. All the best!