Free Expansion, Bulk Modulus and Sound

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Newton's formula for the speed of sound in air gives about two hundred and eighty metres per second. But the real speed is about three hundred and thirty one. Newton, off by fifteen percent? So, what did he miss? On the way, we also meet a gas that expands without cooling. These ideas come up often in JEE, so let's begin.

0:28Free expansion: a gas rushes into a vacuum

So, first, free expansion. Picture an insulated box with a partition. Gas on one side, a vacuum on the other. Pull out the partition, and the gas rushes in; the arrow means goes into. No heat can pass the insulated walls, so Q, the heat, equals zero. Okay? Now tell me. The gas has spread over the whole box. How much work did it do? Yes, exactly, the work is zero! Work needs something to push against, like a piston. In a vacuum, nothing pushes back, so W, the work done by the gas, is zero. Clear? Now the first law. Our sign convention: Q is the heat given to the gas, W the work done by it. Delta, the little triangle, means change in; delta U is the change in internal energy. Delta U equals Q minus W, which is zero. With me? Now, the trap. No heat flows, so this is adiabatic. And adiabatic expansion cools a gas; we saw that in the last video. So this gas cools too, right? No, no, careful here! Delta U is zero, and for an ideal gas, U depends only on the temperature T, okay? The double arrow means so: T does not change. In an adiabatic process, a gas cools only when it does work, and here it does none. Got it? Free expansion is adiabatic, but it is sudden and irreversible. Small p is the pressure, capital V the volume, and gamma is the Greek letter we met in the last video. p V to the power gamma equals constant needs a slow, reversible adiabatic process of an ideal gas, so here it does not hold. Okay?

2:12Bulk modulus of a gas: B=p or B=γp

Now, the bulk modulus: how hard it is to squeeze something. Capital B equals minus V times d p over d V, the rate of change of pressure with volume. Squeeze, and the volume falls while the pressure rises, so the minus sign keeps B positive. Right? First, a slow squeeze at constant temperature: isothermal. From the last video, the slope d p over d V is minus p over V, okay? Put it in, and the V's cancel: B equals p, just the pressure. Now a quick squeeze, with no time for heat to flow: adiabatic. For a reversible adiabatic process, the slope is gamma times steeper, minus gamma p over V, so B equals gamma p. Clear? Gamma is more than one, so the adiabatic bulk modulus is bigger. But why is a gas harder to squeeze when no heat can leave? Yes, exactly! Compress a gas adiabatically, and it also heats up, so its pressure rises even more, and the same fractional squeeze needs more pressure. B ad, with the small letters a d below for adiabatic, is gamma p. That is greater than B iso, with the small word iso for isothermal, which is p; the pointed sign means greater than. Got it?

3:29Speed of sound in a gas: Newton and Laplace

Now, sound. Sound moves through a gas as tiny squeezes and stretches. Its speed, small v, is the square root of B over rho, where rho, the Greek letter, is the density. Stiffer gas, faster sound. Okay? Newton assumed these squeezes happen at constant temperature, so B equals p. That gives v equals the root of p over rho: Newton's formula, and our two hundred and eighty. But think about it. Sound squeezes and stretches the air hundreds, even thousands, of times every second. Is there enough time for heat to flow in and out? No, not enough time! The squeezes are too fast for heat to flow, so they are adiabatic. That was Laplace's correction: B equals gamma p, so v is the root of gamma p over rho. And since p V equals n R T gives p over rho equals R T over M, it is also the root of gamma R T over M, where capital R is the gas constant and capital M the molar mass, okay? Careful with the units. T must be in kelvin, and M in kilograms per mole, the k g mol to the power minus one. So for helium, M is zero point zero zero four, not four. Clear? Divide the two: Laplace's speed is root gamma times Newton's. For air, root one point four is about one point one eight; the wavy equals sign means approximately. So Newton's value is about fifteen percent too low, the gap in our puzzle. Right?

5:01Example 1: gas expands into a vacuum, final T and work done

Alright, the first example. Gas fills one half of an insulated box; the other half is evacuated, which means a vacuum. We remove the partition, and the gas expands into it. It was at three hundred kelvin, the capital K on screen. Find its final temperature and the work done. Pause the video and try it first. Done? Okay, let's check. Some students say: insulated box, so adiabatic, and the volume doubles, so the gas cools, right? No, no, careful, that is the trap! This is free expansion, not a slow, reversible process. Nothing pushes back in a vacuum, so W equals zero. Okay? And the box is insulated, so Q equals zero as well. Clear? So delta U equals Q minus W, which is zero. For an ideal gas, that means the temperature stays the same. With me? So the final temperature is still three hundred kelvin, and the work done is zero. Twice the volume, and not one degree cooler!

6:10Example 2: speed of sound in air at STP, Laplace vs Newton

Now the second example: our puzzle from the start. Find the speed of sound in air at STP, standard temperature and pressure. Density rho: one point two nine three kilograms per cubic metre, the k g m to the power minus three. Pressure: one point zero one three times ten to the power five pascal, the P a, and gamma is one point four. And find Newton's value. Pause the video and try it first. Done? Okay, let's check. For the real speed, which bulk modulus do we use: p, or gamma p? Gamma p, yes, exactly, because sound is adiabatic! So v is the root of gamma p over rho. Put in one point four times one point zero one three times ten to the power five, over one point two nine three, all under the root. Okay? The fraction comes to about one point zero nine seven times ten to the power five. Its square root is about three hundred and thirty one metres per second, the m s to the power minus one. Right? Now Newton's formula: just drop the gamma. The root of one point zero one three times ten to the power five, over one point two nine three: about two hundred and eighty metres per second. That is Newton's number from our puzzle! Quick check: two hundred and eighty times root one point four is about three hundred and thirty one. So, mystery solved! Newton treated sound as isothermal, but it is adiabatic. That missing gamma was the whole fifteen percent. Got it?

7:46Quick revision: free expansion, bulk modulus, sound speed

So, quick revision. Free expansion: Q, W and delta U are all zero, so an ideal gas keeps its temperature. Bulk modulus of a gas: p when isothermal, gamma p when adiabatic. And sound uses the adiabatic one: v is the root of gamma p over rho, or of gamma R T over M, okay? Now go and try the practice problems on this page. All the best!