The Ideal-Gas Equation

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. On a cool morning, a car tyre is at two atmospheres, at twenty seven degrees Celsius. By afternoon, the hot road warms it to forty seven degrees Celsius. Now, forty seven over twenty seven is about one point seven four. So the pressure jumps by seventy four percent, to almost three and a half atmospheres? Something feels wrong here, right? Hold on to that question. Today we learn the ideal-gas equation, which connects the pressure, volume, temperature and amount of a gas. Questions on it come up often in JEE, so let's begin.

0:40One equation for an ideal gas

So, here it is, the ideal-gas equation: p V equals n R T, okay? Small p is the pressure of the gas, capital V its volume, small n the amount of gas in moles, and capital T its temperature. And capital R is a constant, the gas constant. Clear? Its value is eight point three one four joules per mole per kelvin. On screen, J is joules, mol is moles, capital K is kelvin, and each little minus one above means per. Right? Now tell me something. If a gas is at twenty seven degrees Celsius, do we put capital T equal to twenty seven? No, never put twenty seven there! Temperatures here are always in kelvin. The small circle and capital C mean degrees Celsius, and the arrow means becomes. So twenty seven degrees Celsius becomes twenty seven plus two hundred and seventy three, that is three hundred kelvin. Got it?

1:41The same equation in other forms

Now, n, the number of moles, is m over capital M, the mass over the molar mass. It is also capital N over capital N subscript A, the number of molecules over Avogadro's number, six point zero two times ten to the power twenty three per mole. Okay? Now let me ask you this. If we put n equals capital N over capital N subscript A into p V equals n R T, what new constant appears? Yes, exactly! p V equals capital N, times k, times capital T, with k equal to capital R over capital N subscript A, okay? This k is Boltzmann's constant, one point three eight times ten to the power minus twenty three joules per kelvin, the gas constant per molecule. With me? And one more form. Put n equals m over capital M, and bring in the density, mass over volume. We write it as rho, the Greek letter that looks like a p with a tail. We get p equals rho R T over capital M, very handy whenever density appears. Right?

2:52The gas laws hiding inside

Now, the school gas laws all hide inside this one equation. Take a fixed amount of gas at a fixed temperature. What happens to p times V? Yes, it stays constant, because n R T does not change. That is Boyle's law, p V equals const, short for constant. In the graph, pressure p goes up the side and volume V along the bottom. Each curve is for one temperature, so we call it an isotherm, and it is a rectangular hyperbola. Okay? The labels T one, T two and T three have small subscripts, and the less-than signs say T one is the coldest and T three the hottest. So a hotter gas has a higher isotherm. Clear? Next, fix the pressure instead. Then capital V is proportional to capital T, which is Charles's law. This symbol, like an open loop, means proportional to. Got it? Fix the volume, and p is proportional to T, which is Gay-Lussac's law. Keep this one in mind for our tyre puzzle. Right? And Avogadro's law: equal volumes of gases, at the same pressure and temperature, contain equal numbers of molecules, whatever the gas. The double arrow means implies. Okay?

4:07Before and after

Now, a gas in two states, before and after. If the amount is fixed, n R is constant, so p V over T does not change: p one V one over T one equals p two V two over T two. Subscript one means before, and two means after. Clear? But what if some gas leaks out of the cylinder? Can we still use this equation? No, we cannot, because the amount of gas has changed! The same happens when gas is added. So go back to p V equals n R T, and n is proportional to p V over T, okay? Compare n before and after, and you get the fraction that leaked. Right? And one number to remember. At STP, two hundred and seventy three kelvin and one atmosphere, the molar volume is twenty two point four litres. On screen, atm is atmospheres and capital L is litres. Got it?

5:06The trap: absolute pressure

Now, the big trap: always use absolute temperatures and absolute pressures. Say the pressure gauge on a gas cylinder reads two atmospheres. Is the pressure of the gas inside two atmospheres, then? Oh no, be careful here, because the gauge reads only the extra pressure above the one atmosphere outside. So a gauge reading of two atmospheres means an absolute pressure of three atmospheres. With me?

5:31Example 1: tyre pressure from 27°C to 47°C at fixed volume

Now, remember our tyre puzzle? Let's finish it. Example one: a tyre is at two point zero atmospheres at twenty seven degrees Celsius. Find its pressure at forty seven degrees Celsius, if its volume does not change. Pause the video and try it first. Done? Okay, let's check. Here we take two point zero atmospheres as the absolute pressure inside the tyre, not a gauge reading. The volume is fixed and no air leaks out, so p is proportional to T, Gay-Lussac's law. Okay? So the new pressure is two times forty seven over twenty seven, right? No, no, wait, that is exactly the mistake in our puzzle! Those are Celsius readings. Convert first: twenty seven plus two hundred and seventy three is three hundred kelvin, and forty seven plus two hundred and seventy three is three hundred and twenty kelvin. Clear? Now p two equals two times three hundred and twenty over three hundred, which gives two point one three atmospheres. That is a rise of only about seven percent, not seventy four percent. So the driver can relax, and our puzzle is solved.

6:45Example 2: fraction of gas leaked from a cylinder

Now a slightly harder one. Example two: a cylinder holds gas at fifteen atmospheres and twenty seven degrees Celsius. Some gas leaks out, until the pressure is ten atmospheres at seventeen degrees Celsius. What fraction has escaped? Pause the video and try it first. Done? Okay, let's check. The pressure fell from fifteen to ten, so one third of the gas escaped, right? Careful, that forgets the temperature, which also fell! Gas has leaked, so n is proportional to p V over T, and with the volume fixed, n is proportional to p over T, okay? Now in kelvin: twenty seven degrees Celsius is three hundred kelvin, and seventeen degrees Celsius is two hundred and ninety kelvin. Clear? So n two over n one is ten over two hundred and ninety, divided by fifteen over three hundred; each slash means over. That gives zero point six nine zero, the fraction still inside. But the question asks what escaped, not what is left. Right? So the fraction that escaped is one minus zero point six nine zero, which is zero point three one zero, about thirty one percent. Close to one third, but not equal, because the cooling caused part of the pressure drop. Got it?

8:08Quick revision: pV = nRT = NkT, kelvin, absolute pressure

So, quick revision. p V equals n R T, which also equals capital N k T, and p equals rho R T over capital M, with T in kelvin. For a fixed amount of gas, p one V one over T one equals p two V two over T two; when gas leaks, use n proportional to p V over T, okay? And use absolute pressures only: a gauge reading of two atmospheres means three. Now go and try the practice problems on this page. You did really well today. All the best!