Internal Energy, Heat and Work

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Stir some water in a closed flask, hard, for a long time. The water gets slightly warmer, with no flame and nothing hot near it. So tell me, did the water receive heat? Most of us would say yes. But in thermodynamics, the answer is no. Strange, right? Today we make three words completely clear: internal energy, heat and work. Thermodynamics problems in JEE are built on them. So, let's begin.

0:34Internal energy U: a state function

First, internal energy, written capital U on screen. Gas molecules move about randomly, so they have kinetic energy, and they exert forces on each other, so they have potential energy too. The total over all the molecules is the internal energy. Careful: motion of the container as a whole does not count, only the random motion of the molecules. Okay? Now, U is a state function. It depends only on the present state of the system, not on the path it took to get there, just as your height above sea level does not depend on the road you climbed. Right? For an ideal gas, just one variable decides the internal energy. Which one? It is the temperature, yes, exactly! For an ideal gas, U equals f over two, times n R T, okay? Here f is the number of degrees of freedom of a molecule, n the number of moles, R the gas constant, and T the temperature in kelvin. So for a fixed amount of gas, same temperature, same internal energy. Clear?

1:37Heat and work: energy being transferred

Now, heat and work: two ways of transferring energy. Heat, written capital Q, is energy transferred because of a temperature difference. With me? So, back to our stirred water. Was there any temperature difference pushing energy into it? No, there was no temperature difference at all! The energy went in through the moving stirrer, and energy transferred by other means, like a moving piston or a stirrer, is work, written capital W on screen. So the water received work, not heat, and our puzzle is solved! Now a deeper question. Take a gas from one state to another by two different paths: same start, same end. Is the work done the same on both? Not necessarily, and this is the key idea. Q and W are path functions: they depend on the process, not only on the end states. Okay? But the change in U is the same on every path, because U is a state function. So a gas has internal energy, but heat and work are only energy being transferred.

2:45Work done by a gas: the area under the p–V curve

Got it? Now, the work done by a gas. W equals the integral from V one to V two of p, d V, okay? The tall S-shaped sign is the integral sign. V one and V two, with the small subscripts, are the starting and final volumes: V one at the bottom, V two on top. Here p is the pressure of the gas, and d V a tiny change in volume. Right? You have met integrals in maths. So, if we plot p against V, what does this integral give on that graph? The area under the curve, exactly! So W is the area under the p V curve, from V one to V two; the dash means p plotted against V, okay? Topper's trick: often you need no integration at all, just an area. That is also why work is a path function: different paths, different curves, different areas. Clear? Now, in a compression, the volume goes down, so V two is less than V one. What happens to the sign of W? It becomes negative, correct! In an expansion, W is greater than zero: the gas does positive work. In a compression, W is less than zero. Okay? And this is the sign convention the course uses: W is the work done by the gas, and Q is the heat given to the gas. Use it in every problem. Got it?

4:08Work in a cycle: the enclosed area

One more idea: a cycle. The gas comes back to its starting state, so on the p V diagram its path is a closed loop. Okay? The net work in a cycle is the area enclosed by the loop. Here is why: while the gas expands, the area under its path counts as positive work, and while it is compressed, as negative. Subtract, and only the area inside the loop is left. With me? Now, suppose the loop goes clockwise. Then the gas expands along the top, at high pressure, and is compressed along the bottom, at low pressure. So, is the net work positive or negative? Yes, it is positive! The area under the top is bigger than the area under the bottom. So a clockwise loop gives positive net work, and an anticlockwise loop, negative.

4:58Example 1: constant pressure expansion, 10 L to 30 L

Clear? Time for example one. Two moles of a gas expand at a constant pressure of one point zero times ten to the power five pascals, written capital P small a, from ten litres to thirty litres; capital L means litres. Find the work done by the gas. Pause the video, and try it first. Done? Okay, let's check. The problem gives us two moles. Do we need them at all? No, we do not need them! At constant pressure, the path on the p V diagram is a flat line, and the area under it is a rectangle: height p, width V two minus V one. So W equals p times delta V; delta, the little triangle, means change in. Okay? Now, delta V is thirty minus ten, that is twenty litres. So W is ten to the power five, times twenty, right? Check the units first. Oh no, careful here, look at the units! Pascals times litres is not joules. One litre is ten to the power minus three cubic metres, written m cubed, so twenty litres is twenty times ten to the power minus three. With me? Now multiply: ten to the power five, times twenty, times ten to the power minus three. That is two thousand joules, written capital J on screen. Positive, because the gas expanded. Very good!

6:24Example 2: straight-line p-V path, area of a trapezium

Now example two, a little harder. Here, a gas goes from volume two cubic metres and pressure one hundred pascals, to five cubic metres and four hundred pascals; the brackets list the volume first. It moves along a straight line on the p V diagram. Find the work done. Pause the video and try it. Hint: think about the area. Done? Okay, let's check. Many students write W equals p delta V: four hundred times three, twelve hundred joules, or one hundred times three, three hundred joules. So which pressure is right? Neither of them is right, so careful here! The formula p times delta V works only at constant pressure, and here the pressure keeps changing. So use the area under the line, from V equals two to V equals five cubic metres. That shape is a trapezium. Right? Area of a trapezium: half, times the sum of the parallel sides, times the distance between them. Here the parallel sides are one hundred and four hundred pascals, and the width is five minus two, three cubic metres. Okay? Half of five hundred is two hundred and fifty, times three: seven hundred and fifty joules, positive, since the gas expanded. And look, it lies between our two wrong answers, since the pressure stayed between one hundred and four hundred. Perfect!

7:51Quick revision: state vs path functions, work as area

So, quick revision. Internal energy is a state function; for an ideal gas, U equals f over two, times n R T, so it depends on temperature alone. Heat and work are path functions, and the work done by a gas, the integral of p, d V, is the area under the p V curve. In a cycle, the work is the enclosed area, positive for a clockwise loop. Now try the two practice problems on this page. All the best!