WEBVTT
00:00:00.740 --> 00:00:06.420
A cylinder with a movable lid contains gas with a volume of 0.022 cubic meters.
00:00:06.730 --> 00:00:10.080
The gas has an initial temperature of 348 kelvin.
00:00:10.260 --> 00:00:16.950
The temperature is increased to 512 kelvin and the lid rises to equalize the pressure to the value it had before heating.
00:00:17.250 --> 00:00:20.640
What volume does the gas in the cylinder occupy after it is heated?
00:00:20.930 --> 00:00:22.860
Give your answer to three decimal places.
00:00:23.160 --> 00:00:27.860
Let’s say that this is our cylinder containing gas, and we’ve drawn it to show that this lid is movable.
00:00:28.000 --> 00:00:31.030
It can slide up or down to expand or decrease the volume.
00:00:31.350 --> 00:00:35.390
We’re told that the temperature of the gas in this cylinder is increased from its original value.
00:00:35.530 --> 00:00:44.740
This results in an increase in pressure in that volume, which then raises the movable lid so that the final pressure of the gas in this cylinder is equal to the pressure it had initially.
00:00:44.980 --> 00:00:50.840
We’ve seen that this happens through the volume of the gas expanding, and it’s that final volume that we want to solve for.
00:00:51.110 --> 00:00:57.390
If we treat the gas in this cylinder as an ideal gas, then we can describe it using what’s called the ideal gas law.
00:00:57.580 --> 00:01:08.850
In this law, 𝑃 is pressure; 𝑉 is volume; 𝑛 is the number of moles of a gas, that is, its quantity; 𝑅 is a constant value called the gas constant; and 𝑇 is the gas’s temperature.
00:01:09.080 --> 00:01:14.110
It’s fairly common for the quantities of pressure, volume, and temperature to vary in a given scenario.
00:01:14.300 --> 00:01:18.950
It’s less common though for the amount of gas, that is, the number of moles of the gas, to change.
00:01:19.250 --> 00:01:22.050
And 𝑅 we see can’t change because it’s a constant.
00:01:22.270 --> 00:01:26.700
In our situation, the amount of gas, the number of moles of the gas, is a constant.
00:01:26.960 --> 00:01:31.810
That is, the number of particles of this gas is the same, even though the gas expands as it’s heated.
00:01:32.250 --> 00:01:37.590
If we divide both sides of the ideal gas law by the temperature 𝑇, then that factor cancels out on the right.
00:01:37.710 --> 00:01:45.550
What we have then is an equation where on the left there are a set of factors which are fairly likely to change being equal to a set of factors which are not.
00:01:45.900 --> 00:01:49.990
And indeed, we’ve seen that, in our scenario, 𝑛 times 𝑅 is a constant value.
00:01:50.290 --> 00:01:52.470
That has an important implication for us.
00:01:52.590 --> 00:02:12.340
It means if we were to have an initial pressure of our gas, we’ll call it 𝑃 one; an initial gas volume, we’ll call it 𝑉 one; and an initial gas temperature, 𝑇 one, then 𝑃 one times 𝑉 one divided by 𝑇 one would equal a final gas pressure, we’ll call it 𝑃 two, times a final gas volume, we’ll call it 𝑉 two, divided by a final gas temperature, 𝑇 two.
00:02:12.790 --> 00:02:25.450
Because 𝑃 times 𝑉 divided by 𝑇 is equal to a constant, we can say that our gas’s pressure times its volume divided by its temperature at any instant is equal to its pressure times its volume divided by its temperature at any other moment.
00:02:25.910 --> 00:02:32.780
In our problem statement, we’re given some of what we could call initial and final conditions for the pressure, volume, and temperature of our gas.
00:02:33.060 --> 00:02:37.490
For example, at first, the volume of the gas is 0.022 cubic meters.
00:02:37.610 --> 00:02:39.490
We can label that quantity 𝑉 one.
00:02:39.870 --> 00:02:43.650
Likewise, the gas initially has a temperature of 348 kelvin.
00:02:43.870 --> 00:02:45.190
We’ll call that 𝑇 one.
00:02:45.620 --> 00:02:51.090
Then, the gas temperature is increased to 512 kelvin, and we’ll call that 𝑇 two.
00:02:51.400 --> 00:02:55.710
We want to solve for the volume of the gas after it’s been heated, which we’ll call 𝑉 two.
00:02:56.260 --> 00:03:00.220
In this equation, we see that 𝑉 two is present, but it’s not yet isolated.
00:03:00.460 --> 00:03:05.990
We can make 𝑉 two the subject of this equation though by multiplying both sides by 𝑇 two divided by 𝑃 two.
00:03:06.270 --> 00:03:10.350
That way, on the right-hand side, 𝑃 two cancels out as does 𝑇 two.
00:03:10.690 --> 00:03:13.410
That gives us an equation where 𝑉 two is the subject.
00:03:13.600 --> 00:03:20.490
It’s equal to 𝑉 one times the ratio of pressures, 𝑃 one to 𝑃 two, times the ratio of temperatures, 𝑇 two to 𝑇 one.
00:03:20.810 --> 00:03:30.370
Looking back at the information we’ve been given, we see that we’ve been told the two temperatures as well as the initial volume 𝑉 one, but we’re not told anything about the specific values of 𝑃 one and 𝑃 two.
00:03:30.530 --> 00:03:38.470
However, we are told that the lid in our cylinder did rise enough so that the pressure in the cylinder at the end was the same as it was at the beginning.
00:03:38.720 --> 00:03:42.810
That is, the pressures were equalized so that 𝑃 one is equal to 𝑃 two.
00:03:43.050 --> 00:03:48.720
So, whatever is the particular value of these pressures, they’re the same, and therefore their ratio is equal to one.
00:03:49.020 --> 00:03:53.890
Our equation for 𝑉 two then simplifies to 𝑉 one times 𝑇 two divided by 𝑇 one.
00:03:54.100 --> 00:04:02.340
To carry out this final calculation, let’s clear some space at the top of our screen and then substitute in our values for 𝑉 one, 𝑇 two, and 𝑇 one.
00:04:02.800 --> 00:04:05.770
Notice that both of our temperatures are expressed in kelvin.
00:04:05.910 --> 00:04:10.010
Therefore, these units will cancel out, and we’ll be left with final units of meters cubed.
00:04:10.390 --> 00:04:15.190
Rounding this result to three decimal places, we get 0.032 cubic meters.
00:04:15.230 --> 00:04:19.410
This is the final volume of our gas after it’s expanded to equalize pressures.
00:04:19.650 --> 00:04:22.390
And note that this volume is indeed larger than 𝑉 one.