C_v, C_p and Mayer's Relation

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take two identical samples of a gas, and give each the same amount of heat. One is locked in a rigid box, so its volume cannot change. The other sits under a free piston, so its pressure stays fixed. In the box, the temperature rises by forty two kelvin. Under the piston, only thirty kelvin. Same gas, same heat. So where did the missing heat go? Today we meet the two heat capacities of a gas, C v and C p, and Mayer's relation that connects them. This comes up often in JEE, so let's begin.

0:40Two heat capacities of a gas

So, what is a heat capacity? It is the heat needed to raise the temperature by one kelvin, and for a gas we take it per mole. C v, with the small v written below, is the molar heat capacity at constant volume. Here n is the number of moles, d Q over d T is a small heat given over the small rise in temperature it causes, and dividing by n makes it per mole. The capital V below the bracket means the volume is held fixed. Okay? Now C p, with the small p below, is the same, but with the pressure held fixed, like our gas under the free piston. Both are in joules per mole per kelvin. Right? Now think about the rigid box. The volume cannot change. So how much work does the gas do? Zero, exactly! No change in volume, no work, so W, the work done by the gas, equals zero. So in the rigid box, all the heat goes into the internal energy, capital U, okay? That is why C v is also d U over n d T, the rise in U per mole per kelvin. Clear? Now the really useful part. For an ideal gas, U depends on the temperature alone. Delta, the little triangle, means change in. So delta U equals n C v times delta T in every process of an ideal gas, not only at constant volume. Got it?

2:07Mayer's relation: where the extra heat goes

Now, back to our puzzle, with the first law: Q equals delta U plus W, okay? Our sign convention: Q is the heat given to the gas, and W is the work done by the gas. With me? Under the free piston, the gas expands as it warms up. So where else does part of the heat go? Into work, yes! The gas pushes the piston up. At constant pressure, W equals p times delta V, and since p V equals n R T, with capital R the gas constant, that is n R delta T, okay? Now put it all into the first law. The heat, n C p delta T, equals the change in internal energy, n C v delta T, plus the work, n R delta T, right? Now cancel n delta T from every term. What is C p minus C v? It is just R, the gas constant! Eight point three one four joules per mole per kelvin; the minus one powers mean per mole and per kelvin. This is Mayer's relation, per mole of an ideal gas, with the same R for every ideal gas. So C p is always bigger than C v, and the extra R is the work done on the piston. Clear? One more form. Lowercase c p and c v are heat capacities per kilogram, not per mole. Divide by M, the molar mass: lowercase c p minus c v equals R over M, with M in kilograms per mole. Got it?

3:44Gamma, and how the heat is shared

Now, gamma, the Greek letter that looks like a tall y. Gamma equals C p over C v, a pure number with no unit. Okay? Here is a quick question for you. Can gamma ever be less than one? No, it never can. C p equals C v plus R, so C p is always bigger. The double arrow means so: gamma is greater than one for every ideal gas. If you ever get gamma less than one, you have flipped the ratio. Right? Now a topper's trick. Divide Mayer's relation by C v and rearrange: C v equals R over, gamma minus one. And C p, which is gamma times C v, equals gamma R over, gamma minus one. Know gamma, and you know both. With me? Back to the piston once more. At constant pressure, what fraction of the heat actually raises the internal energy? Let's see. Divide delta U, n C v delta T, by the heat, n C p delta T, and the n delta T cancels, so the fraction is C v over C p, that is one over gamma. Clear? And the rest, one minus one over gamma, goes into work on the piston. Got it?

5:05Example 1: cp - cv per kg for nitrogen, M in kg per mol

Alright, time for examples. For nitrogen, with molar mass twenty eight grams per mole, find lowercase c p minus c v, in joules per kilogram per kelvin. Pause the video and try it first. Done? Okay, let's check. The answer is per kilogram, so we use the lowercase form: c p minus c v equals R over M, okay? So we just divide R, eight point three one four, by twenty eight, right? No, no, careful here! That is the trap. The answer is per kilogram, so M must be in kilograms per mole: zero point zero two eight. Divide by twenty eight, and your answer comes out a thousand times too small. Got it? So it is eight point three one four over zero point zero two eight. The wavy equals sign means approximately: about two hundred and ninety seven joules per kilogram per kelvin. Notice: per mole the difference is always R, but per kilogram it depends on the gas. With me?

6:11Example 2: same heat gives 30 K and 42 K rise, find gamma

Now a harder one, and it is our puzzle from the start. The same amount of heat raises the temperature of a gas by thirty kelvin at constant pressure, and by forty two kelvin at constant volume. Find gamma. Pause the video and try it first. Hint: the heat is the same in both cases. Done? Okay, let's check. At constant pressure, Q equals n C p times the rise, thirty kelvin. Clear? At constant volume, Q equals n C v times forty two. Same Q, and the same n, okay? Now, thirty goes with C p, and forty two goes with C v. So gamma, C p over C v, is thirty over forty two, right? Oh no, careful, that is the trap! Equate the two heats: n C p times thirty equals n C v times forty two. The n cancels, so C p over C v is forty two over thirty. The bigger heat capacity gives the smaller rise. And thirty over forty two is less than one, which gamma can never be. Clear? So gamma equals forty two over thirty, that is one point four. With me? And our puzzle is now completely solved. One over gamma is thirty over forty two, that is five over seven. So under the piston, only five sevenths of the heat raised the internal energy; the rest pushed the piston as work. That is where the missing heat went! Got it?

7:47Quick revision: Cp - Cv = R, gamma above 1, 1/gamma into U

So, quick revision. Mayer's relation: C p minus C v equals R per mole of an ideal gas; per kilogram it is R over M, with M in kilograms per mole. Gamma, C p over C v, is always greater than one; C v is R over, gamma minus one, and C p is gamma R over, gamma minus one. At constant pressure, one over gamma of the heat raises the internal energy, and the rest does work. Now go and try the practice problems on this page. All the best!