Isochoric and Isobaric Processes

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Take two identical samples of a gas. One sits in a sealed, rigid box, so its volume cannot change. The other sits under a freely moving piston, so its pressure stays the same. I warm both through the same rise in temperature. Which one needs more heat? Or the same? Hold that answer; by the end, you will know exactly which, and why. This comes up often in JEE, so let's begin.

0:32Two processes, two new words

So, two new words. Isochoric means the volume stays constant, like our sealed box. The Greek letter delta, the little triangle, means change in. So delta V, the change in volume, is zero. Okay? Isobaric means the pressure stays constant, like the gas under the free piston. Capital P, the pressure, stays fixed, while the volume is free to change. Right? Now, one fact from the heat capacity lesson. For an ideal gas, U and H, the internal energy and the enthalpy, depend only on temperature. So in any process, delta U equals n C V delta T, and delta H equals n C P delta T, okay? Here n is the number of moles, delta T the change in temperature, and capital C with a small subscript V or P is the molar heat capacity at constant volume or at constant pressure. Clear?

1:29Isochoric: no work at all

First, constant volume. The walls of the box cannot move. So tell me, how much work is done? Zero, exactly! Pressure volume work needs a change in volume. Delta V is zero, and the double arrow means which gives: small w, the work, equals zero. Okay? Now the first law for a closed system: delta U equals q plus w, where small q is the heat. Sign rule: q is positive when heat is absorbed, w is positive when work is done on the system. With w zero, all the heat becomes internal energy: q with a small subscript V, the heat at constant volume, equals delta U, equals n C V delta T, with me? Now, a trap. No work is done here, so delta H, the change in enthalpy, must be zero too, right? No, no, careful! Delta H equals n C P delta T, and the temperature does change. So delta H is not zero, even with no work done. Many students lose marks right here. Our first example tests this. Got it?

2:37Isobaric: the gas pushes the piston

Now constant pressure. As the gas warms up, it expands and pushes the piston out. So, is work done now? And is w positive or negative? Work is done, and w is negative! At constant pressure, w equals minus P delta V, and for an ideal gas, P V equals n R T, with R the gas constant. So P delta V equals n R delta T, and w equals minus n R delta T: negative on heating, as the gas does work on the surroundings. Okay? Next, the heat: at constant pressure, q with a small subscript P equals delta H, which is n C P delta T, right? And delta U is n C V delta T, as always, because U depends only on temperature. Clear? First law check: q P plus w is n, times C P minus R, times delta T, okay? Since C P minus C V equals R, that is n C V delta T: exactly delta U, so it all fits!

3:43Topper's trick: where does the heat go?

Now, a trick toppers use. When a gas is heated at constant pressure, q, delta U, and minus w in brackets, the work done by the gas, are n delta T times C P, C V and R, so n delta T cancels. Read each colon as is to: q is to delta U is to minus w equals C P is to C V is to R, okay? For a diatomic gas, like nitrogen or oxygen, C V is five over two R, remember? So what is the ratio? Pause the video and think. Seven is to five is to two, very good! C P is C V plus R, seven over two R; double all three. So of every seven parts of heat, five raise the internal energy and two go out as work. Clear? For a monatomic gas, like helium or argon, C V is three over two R and C P is five over two R, so the ratio is five is to three is to two. Example two will confirm it, right?

4:46Example 1: heating 2 mol helium at constant volume

Okay, Example one. Two moles of helium are heated from three hundred kelvin to four hundred kelvin at constant volume. Here mol means moles, and capital K means kelvin. Find q, w, delta U and delta H, okay? Pause the video, and try it first. Done? Okay, let's check. Helium is monatomic, so C V is three over two R, twelve point four seven one, and C P is five over two R, twenty point seven eight five. Capital J is joules, and the little minus ones mean per kelvin, per mole. With me? Delta T is four hundred minus three hundred, one hundred kelvin. And at constant volume, w is zero. Right? So q equals delta U, equals n C V delta T: two times twelve point four seven one times one hundred. That is two thousand four hundred and ninety four joules. Clear? And delta H? No work was done here, so is it zero? Not zero! Remember the trap. Delta H equals n C P delta T: two times twenty point seven eight five times one hundred, which is four thousand one hundred and fifty seven joules. That's it!

6:06Example 2: same heating at constant pressure, q, w, Delta U

Now Example two, a little harder. The same heating, two moles of helium from three hundred to four hundred kelvin; the arrow means to. But now at constant pressure. Find q, w and delta U, okay? Pause and try. Hint: use Example one. Done? Okay, let's check. At constant pressure, q equals delta H, n C P delta T, with the same moles and delta T as Example one. So q is four thousand one hundred and fifty seven joules. Got it? Now the work: w equals minus n R delta T, that is minus two times eight point three one four times one hundred. Minus one thousand six hundred and sixty three joules: negative, as the gas expands and does work. Right? Now, compared with Example one, is delta U bigger, smaller, or the same? The same, exactly! Delta U is n C V delta T, same n, same delta T: two thousand four hundred and ninety four joules. Clear? First law check: q plus w, four thousand one hundred and fifty seven minus one thousand six hundred and sixty three, gives two thousand four hundred and ninety four. Exactly delta U, perfect! And the topper's trick works! q is to delta U is to minus w: these three numbers come to five is to three is to two. The wavy equals sign means approximately. Now our puzzle. The piston needed four thousand one hundred and fifty seven joules, the sealed box only two thousand four hundred and ninety four. The extra heat is exactly the work done pushing the piston out. Right?

7:56Quick revision: isochoric vs isobaric, q : Delta U : -w

So, quick revision. One: at constant volume, w is zero, q V equals delta U equals n C V delta T, and delta H, n C P delta T, is not zero. Two: at constant pressure, w is minus n R delta T, q P equals delta H equals n C P delta T, and delta U is n C V delta T, as always. Three: heating at constant pressure, q is to delta U is to minus w equals C P is to C V is to R: seven, five, two for diatomic, and five, three, two for monatomic. Now try the practice problems on this page. All the best!