Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. An ideal gas rushes into a vacuum: a free expansion. No heat goes in at all, so q is zero. And an entropy change is heat over temperature, so zero over temperature is zero. So the entropy does not change, right? Hmm. But the gas has spread out, more disordered! So which is it? Keep your answer ready; example two will settle it. Today is a concept check class: short why questions from the whole unit. And look, in JEE such why questions come up often. So, let's begin.

0:39Why is work not a property of the gas?

So, first question: why is work not a property of the gas? Look at these three graphs: pressure, capital P, going up, and volume, capital V, going across. In each, the gas expands from V one to V two; the small one and two below are subscripts. Same start, same end. The work done by the gas is the shaded area under its path. So, is the work the same in all three? Not at all! Graph one, a single step against P two, the final pressure: the smallest area, least work. Graph two, two steps: more work. Same start, same end, but three different areas, okay? So the work depends on the path, on how the change is carried out, not only on the start and the end. Real properties, like internal energy, depend only on the state. So work is not a property of the gas. Right? Now graph three: the reversible path, infinitely many tiny steps, and maximum work. Under it, for an ideal gas at constant temperature: n R T, times l n, the natural log, of V two over V one; n is the moles, R the gas constant, T the temperature. So tell me, why does the reversible path give the most work? Think about the push-back. In a reversible expansion, the external pressure, P with the little subscript e x t, equals the gas pressure at every stage. Any harder, and the gas could not expand. Clear? So the opposing pressure is as high as it can be, at every stage, while still allowing expansion. The biggest push-back means the maximum work. Got it?

2:23Isothermal and adiabatic expansion

Here are two reversible expansions of an ideal gas, from the same state, P one, V one, T one, to the same final volume, V two. The blue curve is isothermal: P V equals constant, and T two equals T one. The notes on the right say delta U is zero for it; the little triangle is the capital Greek letter delta, meaning change in, and U is the internal energy. But the molecules are moving apart! So why is delta U zero? Because an ideal gas has no intermolecular forces. No attractions between the molecules, so moving them apart costs no energy at all. Okay? So U depends only on the temperature, capital T, and nothing else. Same temperature, same U: in an isothermal process, delta U equals zero. But only for an ideal gas, remember. Right? Now the orange curve: adiabatic, so no heat flows in or out. Its label says P V to the power gamma is constant; gamma is the Greek letter up top. It falls more steeply, so the area under it, the work done by the gas, is smaller than for the isothermal, as the notes say. And it ends at T three, less than T one: the gas has cooled! The notes say delta U is less than zero. Why? Adiabatic means the heat, small q, is zero. So the first law, delta U equals q plus w, becomes delta U equals w, the work done on the gas. With me? In an expansion, the gas does work on the surroundings, so w is less than zero. No heat comes in to pay for it, so delta U is less than zero: the internal energy falls, and with it the temperature. Clear?

4:06Standard values: enthalpy of formation and entropy

Next, standard values. Delta f H naught: the small f below means formation, and the small circle on top, read naught, means standard. For an element in its reference state, like graphite, or liquid bromine, B r two with a small l in brackets for liquid, delta f H naught is zero. Okay? Now, S naught is the standard entropy. Delta f H naught of graphite is zero, so its S naught must be zero too, right? No, no, careful! First see why delta f H naught is zero. It is only a convention. Enthalpies are measured relative to the elements, so each element in its reference state is simply given zero. Got it? But entropy is different. Entropies are absolute, so S naught of an element at two hundred and ninety eight kelvin is not zero; capital K is kelvin. Clear? Absolute means measured from a true zero. By the third law, S equals zero for a perfect crystal at zero kelvin. So, a convention for enthalpy, a true zero for entropy. This trap comes up often. With me?

5:17Example 1: why an endothermic process can be spontaneous

Now, example one. Why can a process with delta H greater than zero be spontaneous? Delta H, the enthalpy change, is positive here: endothermic, heat is taken in. Pause the video, and try it first. Done? Okay, let's check. Many students say: it takes in heat, so it can never go by itself. Correct? No, no, careful! Energy alone does not decide. At constant temperature and pressure, we judge by delta G, the change in Gibbs energy: delta G equals delta H minus T delta S, where S is the entropy. Spontaneous when delta G is less than zero. Okay? Now, a positive delta H pushes delta G up. But if delta S is positive, the term minus T delta S is negative, and it pulls delta G down. Right? So it is a contest between the two terms. If T delta S is greater than delta H, then delta G is less than zero: spontaneous! Clear? So the gain in entropy drives it, even though heat is taken in. That's it!

6:29Example 2: why entropy rises in a free expansion

Now example two, the harder one, and it is our opening puzzle. Why is delta S greater than zero for a free expansion, even though q equals zero? Pause and try. Hint: entropy is a state function. Done? Okay, let's check. At constant temperature, delta S equals q rev over T: q with the small subscript rev is the heat taken in reversibly. Note the condition: reversibly. Okay? Now, the gas rushes into a vacuum. Is that reversible? No! Nothing pushes back, so the outside pressure differs from the gas pressure by a finite amount: irreversible. So its q equals zero is not q rev, and it cannot go into the formula. That was our puzzle's trap! Got it? Now, here is the key. Entropy is a state function: its change depends only on the start and the end. So pick a reversible path between the same states, and find q rev along it. Right? For an ideal gas, a free expansion keeps the temperature the same: q and w are both zero, so delta U is zero, and delta T is zero. So we choose a reversible isothermal expansion to the same final volume. With me? Along it, the gas does work. And delta U is zero, isothermal and ideal, so the heat absorbed equals the work done. So q rev is greater than zero. Clear? So delta S is greater than zero, even though q is zero. Puzzle solved: the gas spread out, and its entropy really did go up. Perfect!

8:10Quick revision: path functions, ideal gas, standard values

So, quick revision. One: work depends on the path, and reversible expansion gives the most work. Two: for an ideal gas, U depends only on T, so adiabatic expansion cools it. Three: delta f H naught of an element in its reference state is zero only by convention; S naught is absolute. Four: for delta S, always use a reversible path between the same states. Now go and try the practice problems on this page. All the best!