Irreversible Adiabatic Processes and Comparisons

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Two moles of a monatomic ideal gas, at four hundred kelvin and eight bar. It expands adiabatically, so no heat goes in or out, until its pressure is two bar. Done one way, the gas ends at two hundred and eighty kelvin. Done another way, it ends at two hundred and thirty kelvin. Same start, same final pressure, no heat, and still fifty kelvin apart. Strange, right? Hold that thought. This comparison comes up often in JEE, so let's begin.

0:37Adiabatic change against a constant pressure

So, adiabatic means no heat goes in or out, so q, the heat absorbed, is zero. The little triangle is the capital Greek letter delta, meaning change in, so delta U is the change in internal energy. First law: delta U equals q plus w, where w is the work done on the gas. The double arrow means "so": q is zero, so delta U is just w. All the work comes out of the internal energy. Okay? Now, for an ideal gas, delta U equals n C V, times T two minus T one. Here n is the number of moles, and C V, the C with a small V below it, is the molar heat capacity at constant volume. T one and T two are the initial and final temperatures. Right? Now the irreversible part. The outside pressure is suddenly dropped, then held constant while the gas expands. Which formula gives the work now? The constant pressure formula, correct! w equals minus P ext, times V two minus V one, where V one and V two are the initial and final volumes. P with the little subscript e x t is the external pressure. Outside pressure, not gas pressure. With me? But questions usually give pressures, not volumes. So use the ideal gas equation: V equals n R T over P, where R is the gas constant. Clear?

2:04The key equation, and a trap

Now put it all together. Left side: n C V, times T two minus T one. Right side: minus P ext, times the bracket: n R T two over P two, minus n R T one over P one. P one and P two are the initial and final pressures of the gas. This is today's key equation. Write it down, okay? For adiabatic changes, you have seen P V to the power gamma equals constant. Gamma, the Greek letter on screen, is C P over C V, the ratio of the two heat capacities. This change is adiabatic too, so why not just use that? Much quicker, right? No, no, careful! This is the trap. P V to the gamma equals constant holds only for reversible adiabatic changes. Here the gas expands suddenly against a fixed outside pressure: not reversible. Use the key equation instead. Got it?

3:00Reversible against irreversible

Now, back to our puzzle. Same start, same final pressure, and q is zero in both. One expansion is reversible. The other is sudden, against a constant outside pressure. Which one does more work? The reversible one, yes! Reversibly, the gas pushes against almost its own pressure all the way. In the sudden one, it pushes against a low, fixed pressure, so it does less work. And delta U equals w. Less work out means less internal energy lost, so the gas ends warmer. With me? So there's our puzzle, solved! The sudden, irreversible expansion ends at two hundred and eighty kelvin, that capital K on screen. The reversible one does more work and cools further, to two hundred and thirty kelvin. For the same fall in pressure, the reversible expansion cools the gas more. Okay? Now the other direction. If a gas is compressed adiabatically, can its temperature ever fall? No, never! Work is done on the gas, so w is greater than zero, and no heat can leave. So delta U is greater than zero, and the temperature rises. Always. Clear? And a sudden compression to the same final pressure needs more work than the reversible one, so the gas ends even hotter. Keep this pair in mind: irreversible expansion ends warmer, irreversible compression ends hotter. Got it? One special case: free expansion, into a vacuum. There is nothing to push against, so w is zero. With q also zero, delta U is zero, and for an ideal gas, delta T, the change in temperature, is zero too. No work, no cooling at all. Right?

4:47Example 1: final T, 10 bar to 1 bar against constant 1 bar

Now, an example. One mol, that is one mole, of a monatomic gas at three hundred kelvin expands adiabatically against a constant one bar, from ten bar, until its pressure is one bar. Find T two, the final temperature. Pause the video, and try it. Done? Okay, let's check. Monatomic, so C V is three over two times R, okay? The external pressure equals the final pressure, one bar, and P one is ten bar. Right? Now a neat trick: divide the key equation by n R, and both of them cancel. Left: three over two, times T two minus three hundred. Right: P ext over P two is one, so we get T two. And P ext over P one is one over ten, so three hundred becomes thirty. Only ratios of pressures appear, so bar is fine. Okay? Open the brackets: one point five T two, minus four hundred and fifty, equals minus T two, plus thirty. With me? Collect the T two terms: two point five T two equals four hundred and eighty. So T two is one hundred and ninety two kelvin. The gas cooled, as an expansion should. That's it!

6:03Example 2: sudden compression to 5 bar vs reversible value

Now a harder one: a compression. One mole of a diatomic ideal gas at three hundred kelvin and one bar is suddenly compressed by a constant five bar, until its pressure is five bar, with no heat exchange. Find the final temperature, and compare with the reversible value. Given: five to the power two over seven is one point five eight four. Pause and try. Done? Okay, let's check. Diatomic, so C V is five over two times R, okay? P ext equals the final pressure, five bar, and P one is one bar. Clear? Divide by n R, just like before. On the right, last time three hundred became thirty. We divided. So here, three hundred over five, sixty, right? Oh no, careful! That is the trap. Left: five over two, times T two minus three hundred. But on the right, the factor is P ext over P one: five over one, which is five. So three hundred times five: one thousand five hundred. Always write P ext over P one, okay? Open the brackets: two point five T two, minus seven hundred and fifty, equals minus T two, plus one thousand five hundred. So three point five T two equals two thousand two hundred and fifty. Right? T two equals two thousand two hundred and fifty, over three point five. The wavy equals sign means approximately: about six hundred and forty three kelvin. Got it? Now the reversible value. For a reversible adiabatic change, T two equals T one, times P two over P one, raised to the power gamma minus one, over gamma. For a diatomic gas, gamma is seven over five, so that power is two over seven. So, three hundred times five to the power two over seven. With me? Three hundred times one point five eight four is about four hundred and seventy five kelvin. So the sudden compression ends at six hundred and forty three kelvin, against four hundred and seventy five. Much hotter, because it needed more work. Exactly what we said!

8:20Quick revision: key equation, reversible vs irreversible

So, quick revision. One: against a constant external pressure, with q zero, n C V times T two minus T one equals minus P ext, times V two minus V one. Two: P V to the gamma equals constant holds only for reversible adiabatic changes. Three: compared with the reversible change to the same final pressure, an irreversible expansion ends warmer, and an irreversible compression ends hotter. Now try the practice problems on this page. All the best!