Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. One mole of a diatomic gas at three hundred kelvin expands to eight times its volume, and no heat goes in, not one joule. Yet the gas does about three and a half thousand joules of work. So where did that energy come from? We will settle it in example two. Today's mixed set brings Carnot engines, heat capacities and standard processes together, as JEE questions often do. Let's begin.

0:31Idea 1: a Carnot engine and its source

So, idea one, from heat engines. Here it is: a Carnot engine rejects two thirds of the heat it absorbs, to a sink at twenty seven degrees Celsius; the small circle with capital C means degrees Celsius. Find the source temperature. Okay? In the figure, the engine takes heat Q one, the red arrow, from the hot reservoir, the source, at T one: capital Q and capital T, with a small one below. It gives out work W, the orange arrow, equal to Q one minus Q two, and rejects heat Q two to the cold reservoir, the sink, at T two. Eta, the Greek letter, is the efficiency: W over Q one, or one minus Q two over Q one; for a Carnot engine, one minus T two over T one. With me? Put the two together: for a Carnot engine, Q two over Q one equals T two over T one, with absolute temperatures, in kelvin. Right? Now, Q two over Q one is two thirds. So T one is three over two times twenty seven, which is forty point five degrees Celsius. Correct? No, no, careful here, Celsius cannot go into this ratio! Change to kelvin first: twenty seven plus two hundred and seventy three gives three hundred kelvin; capital K stands for kelvin. And Q two over Q one is still two thirds. Clear? So T one is three over two times three hundred: four hundred and fifty kelvin. Got it?

2:06Idea 2: heat capacity in a polytropic process

Idea two, from heat capacities. Find the molar heat capacity of a diatomic gas in the process p V cubed equals constant: small p is the pressure, capital V the volume, and the three up top the power. Okay? Now the figure: p goes up, V across, and four processes start from p naught, V naught; the small zero means the starting value. Flat green line, isobaric; vertical red line, isochoric; and two curves: isothermal, p V constant, and adiabatic, p V to the power gamma constant, with gamma, the Greek letter, equal to C p over C v. Clear? All of these belong to one family, the polytropic process: p V to the power x, constant. Its molar heat capacity, capital C, is C v plus R over one minus x, with C v, small v below, the molar heat capacity at constant volume, and R the gas constant. Got it? Check it on the figure. x equals zero, the flat line: C becomes C p, at constant pressure. x equals one, the isotherm: one minus x is zero, so C goes to infinity, the arrow and the sideways eight. And x equals gamma, the adiabat, with no heat: C is zero. Right? Now, our gas is diatomic. So what is its C v? It is five halves R, yes, for a rigid diatomic gas, as the course takes it. And here, x is three. With me? So C equals five halves R, plus R over one minus three. Watch the sign: one minus three is minus two, so that term is minus R over two. Five halves minus one half: C equals two R, okay? Flip the sign, and you get three R, a common slip. That's it!

4:03Example 1: heat and work for nitrogen at constant pressure

Now, example one. One mole of nitrogen is heated at constant pressure, through twenty kelvin. Find the heat supplied and the work done. Pause the video and try it first. Done? Okay, let's check. First question: which heat capacity do we use here, C v or C p? It is C p, exactly, because the pressure is constant. Nitrogen is diatomic, so C v is five halves R, and by Mayer's relation, C p minus C v equals R, per mole of an ideal gas. So C p is seven halves R, okay? So the heat is Q equals n C p delta T, with n the number of moles, and delta, the little triangle, meaning change in. With R equal to eight point three one four, this chapter's value: seven halves, times eight point three one four, times twenty, about five hundred and eighty two joules; the wavy equals sign means approximately. Right? Now the work done by the gas. At constant pressure it is p times delta V, the change in volume, and using p V equals n R T, this is n R delta T, okay? Eight point three one four times twenty: about one hundred and sixty six joules. Clear? Quick check with the first law, in the course's sign convention: Q is the heat given to the gas, W the work done by the gas, and Q equals delta U plus W, remember? So delta U, the change in internal energy, is Q minus W, about four hundred and sixteen joules, matching n C v delta T, as it should. Perfect!

5:48Example 2: final temperature and work in adiabatic expansion

Now example two, our puzzle from the start. One mole of a diatomic gas at three hundred kelvin expands adiabatically to eight times its volume. Find its final temperature and the work done. Pause the video and try it first. Done? Okay, let's check. Our gas follows the orange adiabatic curve in the figure. But we know the temperature and the volume, not the pressure. So which adiabatic relation links T and V? T V to the power gamma minus one is constant, yes! But this holds only for a reversible, quasi-static adiabatic process of an ideal gas, and the course takes this expansion as reversible. Clear? The gas is diatomic, so gamma is seven fifths, and gamma minus one is zero point four. Got it? Now, the volume becomes eight times. So the new temperature is three hundred times eight to the power zero point four, about six hundred and eighty nine kelvin. Correct? Oh no, careful here, an adiabatic expansion cools the gas! T V to the power zero point four stays constant, so when V grows, T must fall. T two equals T one, times V one over V two, in brackets, to the power zero point four: three hundred times eight to the power minus zero point four. With me? Eight to the power zero point four is about two point three zero, so T two is about one hundred and thirty point six, roughly one hundred and thirty one kelvin. Okay? Now the work. Adiabatic means Q, the heat, is zero. So the first law gives W equals minus delta U, which is n C v, times T one minus T two, in brackets. Right? T one minus T two is about one hundred and sixty nine point four kelvin. So W is five halves, times eight point three one four, times one hundred and sixty nine point four: about three point five two kilojoules. Puzzle solved: with no heat coming in, the gas pays from its internal energy and cools. Perfect!

8:03Quick revision: Carnot, polytropic, isobaric, adiabatic

So, quick revision. Carnot: Q two over Q one equals T two over T one, in kelvin. Polytropic: C equals C v plus R over one minus x, and watch the sign. At constant pressure, the heat is n C p delta T, and the work n R delta T, okay? Reversible adiabatic: T V to the power gamma minus one is constant, and W equals minus delta U, so in an expansion the gas cools. Now go and try the practice problems on this page. All the best!