Heat Capacity and Enthalpy

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Warm some gas by fifty kelvin, twice: once in a sealed can, and once in a cylinder with a free piston, at constant pressure. Same gas, same rise, so the same heat, right? Not so fast! One of them needs more heat. Today we find out which, and where the extra heat goes. This topic comes up often in JEE thermodynamics. So, let's begin.

0:29Heat capacity: heat needed per kelvin

So, heat capacity first. Take a cup of water and a full bucket. To warm each by one kelvin, which needs more heat? The bucket, of course! Heat needed per kelvin is the heat capacity, capital C, right? C equals q over delta T: small q is the heat, and delta T, with the little triangle, the Greek letter delta, is the change in temperature. Unit: joules per kelvin; the minus one power means per. Okay? But that depends on how much water you take. Divide by small n, the number of moles, for the molar heat capacity, C with a small m below it, in joules per kelvin per mole. Divide by small m, the mass, for the specific heat, small c, in joules per gram per kelvin. Clear?

1:14Constant volume or constant pressure?

Now, for a gas, it matters how you heat it. The sealed can is at constant volume, marked by the small V below q and C, okay? The heat equals delta U, the change in internal energy, capital U: q V equals delta U, equals n times C V times delta T, right? For the free piston, the pressure is constant, so a small P appears instead. Now the heat equals delta H, the change in enthalpy, capital H: q P equals delta H, equals n times C P times delta T, okay? So, back to our puzzle. Same gas, same rise. Sealed can, or free piston: which needs more heat? The piston, yes! C P is greater than C V; see the greater than sign. At constant pressure the gas expands as it warms, and pushes the piston out. So part of the heat goes into work of expansion, and you must supply more. Got it? For an ideal gas, the gap is exactly R, the gas constant: C P minus C V equals R, okay? And gamma, the Greek letter like a y, is the ratio C P over C V, clear?

2:25Ideal-gas table: know Cv, then add R

Now the values to know, for ideal gases. Monatomic gases, like helium and argon: C V is three halves R, C P is five halves R, and gamma is one point six seven, okay? Next, diatomic gases: C V is five halves R, and the heading says, know C V, then add R to it. So, what is C P? Seven halves R, correct! Like nitrogen and oxygen, N two and O two; the small two means two atoms. And gamma is seven over five, one point four zero. Right? Last, non-linear triatomic: C V is three R, C P is four R, and gamma is one point three three. So, the topper's trick: remember only C V; add R for C P, then divide for gamma.

3:10Enthalpy: H = U + PV

Clear? Alright, now enthalpy. H equals U plus P V: internal energy, plus pressure times volume, okay? So delta H equals delta U plus delta of P V; the brackets mean the change in the whole product. Now, if the pressure stays constant, what does delta of P V become? P delta V, exactly! Only the volume changes. So at constant pressure, delta H equals delta U plus P delta V, which is q P, the heat at constant pressure. Clear? And for an ideal gas, P V equals n R T, so delta of P V is delta of n R T, where capital T is temperature. Got it?

3:48Ideal gas: U and H depend only on T

Now, an important point. For an ideal gas, U and H depend only on temperature. So, if it is heated at constant volume, is n C P delta T of any use? Yes, it is! Since U and H depend only on temperature, delta U equals n C V delta T, and delta H equals n C P delta T, for any process. Okay? And for an ideal gas in an isothermal process, delta T is zero, so delta U and delta H are both zero. Right?

4:19Example 1: heat for 3 mol N2 at constant P and constant V

Okay, our puzzle with real numbers: Example one. How much heat warms three moles of nitrogen from three hundred to three hundred and fifty kelvin, part a, at constant pressure, and part b, at constant volume? C P and C V are given in joules per kelvin per mole. Pause the video, and try it first. Done? Okay, let's check. Delta T is fifty kelvin. Part a, constant pressure: q P equals n times C P times delta T, so three times twenty nine point one times fifty, four thousand three hundred and sixty five joules. Right? Part b, constant volume: q V equals n times C V times delta T, so three times twenty point eight times fifty, three thousand one hundred and twenty joules. Same rise, but constant pressure needs more heat. Puzzle solved! Clear? So, where does that extra heat go? Into the work of expansion, yes! The extra is four three six five minus three one two zero: one thousand two hundred and forty five joules. Okay? Check with n R delta T: three times eight point three one four times fifty, about one thousand two hundred and forty seven joules; the wavy sign means approximately. Almost exactly the extra; the tiny gap of two joules is only rounding in the given values. It is the work done by the gas in expanding against the constant pressure.

5:50Example 2: q, w, Delta U, Delta H for O2 at constant volume

Now Example two, a little harder. Zero point five zero moles of oxygen, C V equal to five halves R, is heated at constant volume from twenty seven to one hundred and twenty seven degrees Celsius. Find q, w, delta U and delta H; small w is the work. Pause the video, and try it. Done? Okay, let's check. Delta T is one hundred and twenty seven minus twenty seven, one hundred; and a rise of one hundred degrees Celsius is a rise of one hundred kelvin. Right? At constant volume the gas cannot expand, so there is no work of expansion: w equals zero. So q equals delta U, okay? Delta U equals n C V delta T: zero point five times five halves times eight point three one four times one hundred, one thousand and thirty nine joules. So q is the same. With me? The volume is constant, so delta H is the same as delta U, right? No, no, careful! That is the trap. Treat oxygen as an ideal gas: delta H equals n C P delta T for any process, and C P is C V plus R, that is seven halves R here. Zero point five times seven halves times eight point three one four times one hundred: one thousand four hundred and fifty five joules. Clear? Quick check: delta H minus delta U should be n R delta T, okay? That is about four hundred and sixteen joules, and one four five five minus one zero three nine is exactly four hundred and sixteen. Perfect!

7:25Quick revision: qV, qP, Cp - Cv = R and H = U + PV

So, quick revision. One: the heat at constant volume is delta U, at constant pressure it is delta H, and C P is bigger than C V, okay? Two: for an ideal gas, C P minus C V equals R, and C V is three halves R, five halves R or three R, for mono, di and non-linear triatomic gases. Three: H equals U plus P V, and for an ideal gas, delta U and delta H depend only on delta T, on any path. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!