WEBVTT 00:00:01.180 --> 00:00:08.010 A gas with a volume of two cubic meters is initially at a temperature of 300 kelvin and a pressure of 500 pascals. 00:00:08.420 --> 00:00:12.920 It is heated to 375 kelvin while being kept at a constant volume. 00:00:13.680 --> 00:00:16.610 What is the pressure of the gas after it has been heated? 00:00:17.510 --> 00:00:20.550 This question is about a gas that’s kept at a constant volume. 00:00:20.580 --> 00:00:24.330 So, we could imagine it as a gas inside a box like this with a fixed volume. 00:00:25.150 --> 00:00:27.910 A gas consists of particles that are free to move around. 00:00:28.050 --> 00:00:33.270 And so we’ve got all these particles, that’s the pink circles in our diagram, flying around inside the box. 00:00:33.830 --> 00:00:37.880 Here, we’ve indicated the direction of each particle’s motion with an orange arrow. 00:00:38.660 --> 00:00:44.080 Those particles can collide with each other, and also importantly they can collide with the walls of the box. 00:00:44.750 --> 00:00:50.120 Let’s consider a particle colliding with one of the walls, for example, like this one right here is about to do. 00:00:50.770 --> 00:00:55.380 Remember that we’re told that the volume is constant, which means that the walls of the box are fixed. 00:00:55.610 --> 00:00:59.510 This means that when the particle collides with the wall, then it will simply bounce off. 00:00:59.780 --> 00:01:01.960 This process exerts a force on the wall. 00:01:02.730 --> 00:01:07.120 This force is always going to have a component that’s acting outward perpendicular to the wall. 00:01:07.800 --> 00:01:19.490 Since in reality there’s going to be a much, much larger number of particles flying around than we’ve drawn in this sketch, then that means that there’s continually going to be a large number of particles colliding with and bouncing off the walls of the box. 00:01:19.570 --> 00:01:23.220 So, that’s a large number of collisions happening across the area of the walls. 00:01:23.520 --> 00:01:32.000 Since each of these collisions will exert a force on the wall with an outward component, then that means that there’s a force acting all over the area of the walls of the box. 00:01:32.450 --> 00:01:36.350 A force acting across a surface with a particular area results in a pressure. 00:01:36.490 --> 00:01:39.320 And so there’s a pressure being exerted on the walls of the box. 00:01:40.120 --> 00:01:47.930 In this question, we’re told that this pressure is initially equal to 500 pascals and that this is the case when the temperature of the gas is 300 kelvin. 00:01:49.050 --> 00:01:53.250 Let’s label the initial pressure as 𝑃 one and the initial temperature as 𝑇 one. 00:01:53.700 --> 00:01:58.130 We’re told that the gas is then heated to a temperature of 375 kelvin. 00:01:58.740 --> 00:02:01.120 We’ll label this new temperature as 𝑇 two. 00:02:01.330 --> 00:02:05.360 Let’s also label this new pressure at the temperature 𝑇 two as 𝑃 two. 00:02:05.580 --> 00:02:07.920 And this pressure is what we’re being asked to find. 00:02:08.410 --> 00:02:16.530 To understand why the pressure of this gas will change when the temperature changes, we can recall that temperature is related to the average speed of particles in a gas. 00:02:16.720 --> 00:02:21.050 And in particular the higher this temperature, the greater the average speed of the particles. 00:02:21.710 --> 00:02:32.030 Now, if we have a box of gas like this with a constant volume and we imagine a particle colliding with one of the walls, then the faster that particle is moving, the greater the force it will exert on the wall. 00:02:32.950 --> 00:02:40.820 If we have many particles moving with a greater average speed, then on average each particle will exert a greater force on the wall when it collides with it. 00:02:41.530 --> 00:02:45.610 This means that a greater average speed results in a greater value of pressure. 00:02:46.250 --> 00:02:53.580 Since we know that the average speed indicates the temperature of the gas, then this means that a higher temperature results in a greater pressure. 00:02:54.090 --> 00:03:00.660 This is described by our law known as Gay-Lussac’s law, which says that pressure 𝑃 is directly proportional to temperature 𝑇. 00:03:01.480 --> 00:03:05.650 This can also be written as 𝑃 is equal to a constant multiplied by 𝑇. 00:03:06.370 --> 00:03:10.850 It’s important to note that Gay-Lussac’s law applies for gas held at a constant volume. 00:03:11.280 --> 00:03:17.540 Since the question tells us that this gas is kept at a constant volume while it’s heated, then we know that this formula will apply. 00:03:18.200 --> 00:03:26.470 If we divide both sides of the equation by the temperature 𝑇, then on the right-hand side, the 𝑇 in the numerator will cancel with the 𝑇 in the denominator. 00:03:27.260 --> 00:03:31.950 This leaves us with an equation that says 𝑃 divided by 𝑇 is equal to a constant. 00:03:32.490 --> 00:03:40.150 So that means that for a given gas kept at a fixed volume, the pressure of that gas divided by its temperature will always have the same constant value. 00:03:40.970 --> 00:03:49.660 In this question, we know that at some instant initially the gas has a pressure of 𝑃 one equal to 500 pascals and a temperature of 𝑇 one equal to 300 kelvin. 00:03:50.420 --> 00:03:57.110 Then, after it’s heated, it has a temperature 𝑇 two of 375 kelvin and a pressure that we’ve labeled as 𝑃 two. 00:03:57.870 --> 00:04:07.670 Since Gay-Lussac’s law tells us that the pressure divided by the temperature is constant, then this means that 𝑃 one divided by 𝑇 one must be equal to 𝑃 two divided by 𝑇 two. 00:04:08.470 --> 00:04:11.430 The quantity that we’re trying to find is this pressure 𝑃 two. 00:04:11.670 --> 00:04:17.210 And we can make 𝑃 two the subject of the equation if we multiply both sides by the temperature 𝑇 two. 00:04:17.790 --> 00:04:21.460 On the right, the 𝑇 two in the numerator and denominator cancel out. 00:04:21.920 --> 00:04:28.740 Then, writing the equation the other way around, we have that 𝑃 two is equal to 𝑃 one times 𝑇 two divided by 𝑇 one. 00:04:29.350 --> 00:04:35.630 Let’s now substitute our values for the pressure 𝑃 one and the temperatures 𝑇 one and 𝑇 two into this equation. 00:04:36.230 --> 00:04:44.170 We find that the pressure 𝑃 two is equal to 500 pascals multiplied by 375 kelvin divided by 300 kelvin. 00:04:44.860 --> 00:04:50.310 The units of kelvin cancel from the numerator and denominator, which leaves us with units of pascals. 00:04:50.860 --> 00:04:55.850 Evaluating the expression then gives a result for 𝑃 two of 625 pascals. 00:04:56.500 --> 00:05:02.820 So then we have found that the pressure of the gas after it has been heated is equal to 625 pascals.