Heat, Work and the First Law

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Open a physics book, and the first law of thermodynamics often reads: delta U equals q minus W, minus! Open your chemistry book: delta U equals q plus w, plus! Same law, opposite signs. So one of the two books must be wrong, right? Hmm. By the end of this lesson, you will answer that yourself. In JEE, questions on heat and work come up often, and one wrong sign costs you the whole answer. So, let's begin.

0:32The first law: energy is conserved

So, let's build the law. Take a closed system, say some gas in a cylinder. Energy can cross its boundary, but matter cannot. Now tell me, in what ways can energy get into this system? Yes, exactly! As heat, or as work. Those are the only two ways in. Okay? And energy can be neither created nor destroyed. That is the first law of thermodynamics. Right? So, for a closed system, delta U equals q plus w. The little triangle is the capital Greek letter delta, and it means change in. Capital U is the internal energy of the system, small q is the heat, and small w is the work. The internal energy changes by exactly what comes in. Clear?

1:20Signs: look from inside the system

Now, the most important part of today: the signs. This course follows the IUPAC convention. Imagine you are sitting inside the system. Heat comes in, it is absorbed, and your energy goes up. So q is greater than zero; that is the greater-than sign on screen. Positive. Okay? Same for work. Push a piston in, and work is done on the system: energy goes in. So w is greater than zero, positive as well. Right? Now you tell me. The gas pushes the piston out, so the work is done by the system. Plus or minus? Minus, correct! Energy leaves the system, so it is negative. And heat given out by the system? Negative too. So the rule is really simple: into the system, plus; out of it, minus. With me?

2:09Work against a constant external pressure

Now, how do we calculate the work? In chemistry, it is mostly pressure volume work: a gas pushing against the outside. When the external pressure stays constant, w equals minus P ext times delta V, with a minus sign in front. P with the little subscript e x t is the external pressure, and delta V is the change in volume. Clear? But why this minus sign? When a gas expands, delta V is positive. And who is doing the work then? The gas, yes! The system does the work, so by our convention, w must be negative. That is exactly why the minus is there. The double arrow on screen means which gives. With me? And compression is the opposite. The volume shrinks, so delta V is negative, and w comes out positive: work is done on the gas. So put in delta V with its sign, and the formula does the rest. Got it? One more thing: units. Pressure in atmospheres times volume in litres gives litre atmospheres, written L atm. One litre atmosphere is one hundred and one point three joules; the capital J means joules. And one litre bar is exactly one hundred joules. Our second example needs one of these. Okay?

3:26Three special cases

Alright, three special cases that make questions quick. First, constant volume, like a rigid sealed vessel: no pressure volume work, and with no other kind of work, w is zero. Then delta U equals q, written q with a subscript V, the heat at constant volume. Right? Second, adiabatic: no heat crosses the boundary, like gas in an insulated cylinder. So q is zero, and delta U equals w. Compress that gas, w is positive, and its internal energy goes up. With me? Third, a cycle: the system comes back to exactly the state it started from. Remember, U is a state function. So after one full cycle, what is delta U? Zero, very good! Then zero equals q plus w, so q equals minus w. Whatever net heat goes in comes out as work. Clear?

4:21Physics books: the same law, another sign

Now, back to our puzzle. In chemistry, delta U equals q plus w, where small w is the work done on the system. Okay? In many physics books, delta U equals q minus capital W, where capital W is the work done by the system. Work done by the system is just minus the work done on it. So the minus sign has simply moved into the definition. Neither book is wrong! Let's test it. Here, a gas absorbs five hundred joules of heat and expands, doing two hundred joules of work. Find delta U both ways. Physics: five hundred minus two hundred. Chemistry: q is plus five hundred and w is minus two hundred, so five hundred plus minus two hundred. Plus three hundred joules in both! Just never mix the two. Got it?

5:12Example 1: Delta U, 250 J heat given out, 400 J work done on

Okay, Example one. Here, a system gives out two hundred and fifty joules of heat, while four hundred joules of work is done on it. Find delta U for the system. Pause the video, and try it first. Done? Okay, let's check. Two fifty and four hundred. Just add them: six hundred and fifty joules, right? Oh no, careful! Read the words: the heat is given out. Energy leaves the system, so q is minus two hundred and fifty joules. Okay? And the work is done on the system, so energy comes in. So w is plus four hundred joules. Right? Now, delta U equals q plus w: minus two fifty plus four hundred. Plus one hundred and fifty joules. More energy came in as work than went out as heat, so the internal energy goes up. That's it!

6:05Example 2: gas expands 250 cm3 against 1 atm, find Delta U

Now Example two, a little harder. Here, a gas absorbs forty joules of heat, and expands by two hundred and fifty c m cubed, that is cubic centimetres, against one atmosphere; a t m on screen means atmosphere. Find delta U for the gas. Pause and try. Hint: watch the units. Done? Okay, let's check. Heat is absorbed, so q is plus forty joules. Easy, okay? Now the work. The external pressure stays at one atmosphere, so w is minus P ext delta V, with P ext equal to one. Minus one times two hundred and fifty: minus two hundred and fifty joules, right? No, no, wait! Units! Atmospheres times cubic centimetres is not joules. Change the volume to litres; capital L means litres. One litre is a thousand cubic centimetres, so delta V is zero point two five zero litres. With me? Now, minus one atmosphere times zero point two five zero litres: minus zero point two five zero litre atmospheres. Negative, because the gas expands and does the work. Clear? Now convert. One litre atmosphere is one hundred and one point three joules. So w is minus twenty five point three joules. Okay? Finally, delta U equals q plus w: forty minus twenty five point three. Plus fourteen point seven joules! Forty joules came in as heat, about twenty five went out as work, and the gas kept the rest. The signs tell the whole story!

7:38Quick revision: sign convention, w = -P Delta V, cases

So, quick revision. One: delta U equals q plus w, and into the system is plus, out of the system is minus. Two: against a constant external pressure, w equals minus P ext delta V, and one litre atmosphere is one hundred and one point three joules. Three: at constant volume, delta U equals q V; adiabatic, delta U equals w; and round a cycle, q equals minus w. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!