00:10 Good day everyone, my name is Adekunle, I'll be teaching you chemistry and the topic for today

00:18 is gas loss. The first gas loss we are going to talk about is boy's law. According to the

00:25 about board in 1662, it says the volume of the fixed mass of gas is invested proportional

00:31 to its pressure provided that the temperature remains constant. I would say V,

00:35 proportional, inverse proportional to P, V, as you can see from the board. V representing volume

00:44 P, representing pressure, we obtain P1, V1 equals to P2, V2. And according to the candidate theory,

00:52 the gas pressure is caused by molecular collisions with the walls of the container. Therefore,

00:58 the larger the normal of molecules per unit volume, the larger the normal of collisions,

01:02 and the higher the pressure, which implies that the more the pressure, the lesser the volume,

01:07 and the more the volume, the lesser the pressure. For example, 375 cm cube of gas

01:15 asks a pressure of 70 mm mercury. Find its volume in the pressure is reduced to 750 mm mercury.

01:25 How do we solve this? You are given the volume, 375 cm cube. And the pressure at that volume is

01:37 77 mm mercury. Find its volume, V2. Find its volume. Question mark, if the pressure is reduced to

01:47 is reduced to 750 mm mercury. If you apply Boyz Law that says P1, V1 equals to P2, V2.

01:59 And making V2 the subject implies that you divide both sides by P2. If you do that,

02:05 if you do that, you will obtain V2 to be equals to P1, V1, all over P2. Putting the values you have

02:16 on your board, you discover V2 equals to 77C times V1, 375, all over P2, which is 750. If you point

02:30 to concrete, very well, you obtain 385 cm cube. I believe you understand. And then the activity is

02:39 plot the graphical presentation of Boyz Law forming. Chast Law is the next law.

02:45 This shows the relationship between volume and temperature. And Chast Law states that the volume

02:50 of the giving mass of gas is directly proportional to its absolute temperature in KV,

02:54 where the pressure remains constant. How do we show this? This tells us that V is

03:03 proportional to temperature. If you follow due process of combating, putting your

03:11 proportional to constant, you obtain at the end of the V1, all over T1 equals to V2, all over T2.

03:21 And according to the connaissance theory, what the test means is that the higher the volume,

03:26 the higher the temperature. E volume increase, temperature we also increase,

03:31 provided temperature remains constant. And at KV temperature, which is absolute temperature,

03:35 we are giving it to be at -273 degrees Celsius. And the conversion from degree Celsius to Kelvin

03:44 simply says at one Kelvin, you are going to have taking this to the other side, you are going to have

03:54 -272 degrees Celsius. So at the end of the day, temperature in KV equals to degree Celsius plus 273.

04:02 Or temperature in degrees Celsius equals to temperature in KV minus 273. For example,

04:09 combined following from Celsius temperature to Kelvin temperature, I'll just take the first one,

04:14 which says 143 degrees Celsius. Temperature in Kelvin equals to temperature in degrees Celsius plus

04:22 273. As I've seen before this present slide, then giving that temperature in degrees Celsius is 148

04:31 degrees Celsius, then you are 273. If you do that, you obtain your value in Kelvin. And what do

04:39 you get? 4 on rate and 21 Kelvin. With this, you should be able to do zero degrees Celsius.

04:45 Downing is automatic. It's just going to give you 273 Kelvin. And minus

04:49 132 degrees Celsius. Just at 273, you obtain 141 Kelvin. And an example,

04:58 the real example on child's law. Say the sample of gas has a volume of 125 CM cube at 17 degrees

05:06 Celsius, what will be the volume at 100 degrees Celsius if the pressure remains constant? Volume,

05:12 you're told 125 CM cube. At 17 degrees Celsius,

05:19 combined to Kelvin, which implies you are 273. And you obtain what? 290 Kelvin. You have to find

05:29 the volume at this temperature on the degrees Celsius. Adding to 73, you obtain what? 373.

05:37 Adding to 73, you obtain 373 Kelvin. And to find V2, from child's law, we are told that V1

05:45 over T1 equals to V2 all over T2. Making V2 so that's what it means to take this T2 to the other side.

05:52 You're going to obtain V2 equals to V1 T2 all over T1. Then putting the values

06:00 on the board, you have V1, 1, 2, 5, times T2, 373, all over V, all over T1, which is 290.

06:12 If you point your calculator very well, you obtain 161 CM cube. I believe you understand that.

06:22 The general gas equation. The general gas equation. Combining voicelo and chastelo,

06:29 you obtain a joint variation which says V, proportional to T, proportional to 1 over P.

06:34 Making change in the proportionality to putting a proportionality constant, you remove the

06:42 proportionality and it puts equation. At the other day, you obtain K to be equals to Vp over T,

06:48 which at the end of the day, you are going to obtain P1, V1, all over T1 equals to P2, V2 over T2.

06:59 The general gas equation, which is on the board already, you have seen that.

07:04 And then, because we will need standard temperature and pressure, which is given at 273kV

07:09 and for pressure, it is given at 7.60 minutes of recovery, or 1 at 2.3, or 1.01, 10.45,

07:16 depending on the values you are giving. For example, it gives a mass of gas, it gives a mass of gas

07:22 occupies 8.50 CM cube at 3.20 KV and 0.92 times 6.5 Nm square pressure calculates the volume.

07:31 Calculates the volume of the gas at STP. Calculate the volume of the gas at STP.

07:42 You are giving V1 at 50 CM cube at this temperature, 3.20 KV. And this particular pressure,

07:54 having these units, Newton per meter square, you can guess the corresponding pressure,

08:04 because we are told at STP, corresponding pressure for Newton per meter square is 1.01,

08:10 10.45 Newton per meter square. And then, the temperature at that standard temperature

08:17 pressure is 2.73 KV. You are asked to find V2. Making V2 the subject of formula from the idea,

08:27 from general gas equation, you obtain V1, V1, T2 all over T1, P2. And put the values in them,

08:38 you are going to obtain at the end of the day, C60.5 CM cube. I believe you can do that

08:46 from the previous examples we've been doing so far. You are not a JJC anymore.

08:51 And now, with students, I believe you are following idea gas equation. In all experiments,

08:57 our four quantities are very important. What are these quantities? Volume, pressure,

09:02 temperature, and non-ball of moves, or mass. And then, combined with Avogado's flow,

09:08 we say the equivalent of four gases at the same temperature and pressure,

09:11 continue the same non-ball of molecules. Also, taking note of 1 mole of any gas at STP,

09:16 which is 22.4 DM cube. Combining all this together with boys and chastelo,

09:21 you obtain the idea gas equation. We say SPV, because of LC, for 1 mole of gas,

09:26 and for M mole of gas, you are PV, because of NRC. We have P is in atmosphere, V in Q, C in Kevin.

09:33 If you do that, you are going to obtain the value of R, you are going to obtain the value of R

09:37 in this way, PV over NC, P1 atmosphere, and volume 22.4 DM cube,

09:49 all of the non-ball of moves, 1, and temperature 273. If you point your capitope row,

09:56 you obtain the molar gas constant R at 0.082 atmosphere, DM cube,

10:03 Pachevin, Pamo, which is the value I've seen on the board. An example,

10:11 a certain amount of gas occupies 5.0 DM cube at 2 atmosphere and 10 degrees Celsius,

10:17 calculates the normal of moves present, calculates the normal of moves present,

10:22 the value of R is intact as concealed in the board, and then you are told that the volume is

10:27 5.0 DM cube at this pressure, and this temperature, temperature must be in degrees sessions,

10:39 in Kelvin's story, you are 273. So obtain 283 Kelvin. Put all these values into our

10:51 I do gas equation. A making end is object of formula, you obtain PV over R, C.

11:00 I put all these values there. P is 2, volume is 5, R is 0.0, H2 times temperature 283.

11:13 If you point your capitope row, you obtain this value for non-ball of moves. You obtain these values

11:21 for non-ball of moves. I believe you have followed this very well. Then you should be able to

11:26 understand the next examples. Since 1.5 more of an ideal gas at the temperature of minus 15 degrees

11:34 Celsius, and pressure of 2 atmosphere, what volume in DM cube really gas occupies at that temperature?

11:41 Given the value of the molar gas constants R to be 0.082 atmosphere DM cube per more per Kelvin,

11:48 how do you solve this? From the previous examples you should be able to do this,

11:53 L is given as 1.5 moves. Temperature to be combated to Kelvin, minus 15 degrees Celsius,

11:58 plus 173. You obtain your 258 Kelvin, and the pressure of 2 atmosphere are 0.082.

12:04 Put in the ideal gas equation there, PV because of NRT, and making V the subject of formula,

12:11 you discover that NRT over P, then put the values in, you obtain 1.5 times 0.082 times 2.5

12:19 all over 2 atmosphere. You obtain the value of volume to be 15.87. Then the next

12:26 is doubting slow of pressure pressure. This is that if there is a ratio of gases, we do know that

12:32 chemically together, can we do that pressure? Is that about the ratio? Is it some of the individual

12:37 pressure that make up the ratio? In the individual pressure pressure, that make up the ratio P to

12:41 take also P, A plus P, B plus PC, to PN, to the number of pressure, to the number of pressure pressures

12:48 you are giving. For example, you are told a certain mass of hydrogen gas collected over water

12:55 at 6 degrees Celsius and 765 millimeter mercury pressure. Ask a volume of 35 cm cube, calculate

13:03 the volume when it is dry at SCP. If you observe this question very well, you have to calculate

13:10 the volume of the gas. Here, you are told that the gas is collected over water. You don't need that

13:18 over water, but that is why you are giving saturated vapor pressure over water at 6 degrees Celsius

13:24 to be 7 mm mercury. The pressure of the gas is actually P water vapor,

13:40 plus pressure plus the total pressure. The total pressure is the pressure of the gas plus

13:50 pressure of the water vapor. And to make the pressure of the gas, the sublet of formula,

13:57 you are by taking this to the other side, you have P to the minus P water vapor.

14:06 P to the as you are giving is 765, 765 minus the total vapor pressure of water at 6 degrees Celsius,

14:16 which you are told that it is 7 mm mercury. If we subtract this, you have the pressure of the gas

14:22 to be 758. And that represents our P1, because we are going to use the general gas equation.

14:33 And you are told from the question that V1 is 35 cm cube. The temperature T1, which is 6 degrees

14:43 Celsius, to be combated to Kelvin, you are 273, then you obtain 279 Kelvin. P2 from the question

14:55 at standard temperature of pressure is 76 mm mercury. And V2 is what we are looking for. T2 is

15:07 temperature at standard. Temperature at standard, which is 273 Kelvin. To solve this,

15:16 apply the general gas equation, apply the general gas equation, P1, V1, all over T1,

15:23 because to P2, V2, all over T2, making V2 the subject of formula T2 recommended, P2 recommended,

15:32 so we obtain V2 equals P1, V1, T2, all over P2 over T1, P2 over T1, then V2, putting all the values in,

15:47 you have 758 times 35 times 273, all over 76, and then T1 is 279. You obtain, at the end of the

16:05 if you put your capitol very well, 34.2 cm cube, 34.2 cm cube. The next law we talk about is

16:15 Galusas law of combined in volume. This is the way gases react. They do so in volume, which are

16:22 in simple ratios, one another, and to the volume of the produce, if gaseous, provided at temperature

16:27 and pressure remains constant. For example, one volume of oxygen is required for the complete

16:34 combustion of 15 cm cube, of 15 cm cube, of 18. What's formula for 18? 18 is C2, H6. If all gases

16:45 are measured at STP, combustion of 18, C2, H6 plus O2, certainly you are going to have

16:54 our product to be carbon four inside, and what's that? With our ratio to most time,

16:59 less balance seats, you have two ears, you have seven ears, you have four, and then you have six.

17:08 And if we try the question, you are comparing 18 and oxygen, and from the balanced equation,

17:14 you have two volume of 18 to react with seven volume of oxygen. And you are asked that

17:29 what volume of oxygen, what volume of oxygen is required for the complete combustion of 15

17:38 volume of this? 15 volume of this, and what are you going to have cross multiply to give you two times

17:45 X, two X, and then 15 times seven. The bubbles are by two. Your S, we certainly give you 52.5 cm cube,

17:59 of oxygen. I believe you understand that. For marbogadro's law, this is the next law. This is

18:06 that equal volume of gases at the same temperature and temperature, contain the same number of molecules.

18:12 Given the, a particular reaction of hydrogen and oxygen to give us water,

18:18 if you compare Avogadro's law, and give us our law, give us our thoughts about the volume,

18:23 why Avogadro's law talks about the molecules? Let's compare more Avogadro's number,

18:29 amolavoli, what relation is she? Fine. A mole of any softness is the amount which it contains as many

18:36 elementary particles as there are in 12 grams of carbon 12. The normal of specified particles

18:42 present in one mole of any softness is equal to 6.02. 6.02 times 6.33. This value is known as water,

18:49 Avogadro's constant. Avogadro's constant. Let's compute number of moles.

18:56 Number of moles is equal to mass over molar mass.

19:04 Number of moles is also equal to volume over molar gas volume.

19:16 And number of moles from Avogadro's number to use Avogadro's number,

19:21 I thought you were going to compare relationship between moles Avogadro's number and molar volume.

19:25 Normal of moles equals to number of particles or atoms or even molecules

19:35 over Avogadro's number. 6.02 times 10 is the power 23.

19:43 If I want to add the molar volume, number of moles equals to, okay, it's already here,

19:50 equals to volume and the molar gas volume is 22.4 DMQ.

19:58 This is what I've shown on the board already. From Avogadro's low,

20:04 one mole of any gas at STP as the volume of 22.4 DMQ. This value is known as water,

20:10 mole of gases. The mass low we talk about is grams low of diffusion according to grams.

20:20 It says that at a particular temperature and pressure, the rate of the evolution of the gas is

20:25 inversely proportional to the square root of its density. This is also, it can also be stated

20:31 that at a particular temperature and pressure, the rate of the evolution of the gas is

20:34 inversely proportional to its relative molecular mass. As I'm going to see in the next example,

20:40 we are going to talk about. I would like to show this. Rate of the evolution is inversely

20:47 proportional to the square root of its density, this is representing density rho.

20:51 And as a good commissive student and mathematicians, you put equations and then

21:01 wouldn't you put an additional constant K? You obtain the square root over the square root of

21:07 density. Again, I said this is rho and it's representing density.

21:14 If you cross multiply, you're going to have rate,

21:26 it also constant means that as long as you have many rates of diffusion and as long as

21:32 there are many densities, you can't be comparing them. And then to represent this time that you are

21:38 going to have R1 over R2 equals the square root of density 2 over density 1. This is grams low

21:47 of diffusion. And like I told you before, that it can also be stated as inversely proportional

21:53 to the square root of its relative molecular mass. If you do that, you're still going to obtain

21:58 something. This is representing relative molecular mass. That's what we are seeing on the board.

22:05 From grams low of diffusion, we have this example. If 280 cm cube of hydrogen diffuse in 40 seconds,

22:12 how long will it take for 490 cm cube of a gas? Yes, whose vapor density is 25 to diffuse

22:19 on that same condition? Given the relative molecular mass of hydrogen gas to be 2.

22:24 First, the volume of hydrogen, volume of hydrogen, is given as 280 cm cube.

22:42 Diffuse does time diffuse in 40 seconds does the time 40 seconds. How long will it take? That's

22:51 another time for that gas x. This is what I must find. And I'm giving the volume of gas s to be 490

22:59 cm cube. And giving the vapor density also of this gas s to be 25 automatically,

23:08 the relative molecular mass of s is 2 times vapor density, which is 2 times 25. And this give us 50.

23:17 What is are you giving? relative molecular mass of hydrogen as 2. From grams low of diffusion,

23:26 R1 over R2 is equal to square root of relative molecular mass of 2 over relative molecular mass

23:38 of 1. Now comparing with hydrogen and the gas s, R8 over Rs equals to square root of

23:50 that m represents our relative molecular mass this time over these, put in the values. And then

24:00 Gavin's knowledge that rate is volume over time. Our simple applied that's here also,

24:09 that mean rate of h is volume of h over time h. Also, over volume of s over time s equals to square

24:23 root of rs over relative molecular mass of hydrogen. Put in the values in. Put in the values in

24:36 volume of hydrogen is 280, is 280. Over time of hydrogen is 40 seconds. Over volume of x is 490.

24:53 Over time of s, I'm looking for it. Let the molecular mass of s you are giving,

25:00 or we calculate it, we obtain 50 all over the one of hydrogen is 2. Here 1, here 25,

25:08 you obtain square root of 25. And square root of 25 is 5, you know that. Here we already have 270 over 4,

25:18 280 over 4th, over 4th is sorry, 4, here 2 and then 14, 2, here 1, 2, here 7. We put the values

25:29 of vh and over th, all over vs over th equals to this. All the values where it's stable, we

25:37 have to build it before, then we just put them in. You have 280 over 40, over all over 490,

25:43 over ts equals to square root of 50 over 2. Court 2 goes here 1, 2 in 50, 25, then you obtain square root

25:51 of 25, that will give us 5. And on this side 280 over 40, we get 7, 7 over 490, over tx.

26:08 How do we go about solving this? Cross multiply, you have 7, you have 7 equals 5 times 490,

26:19 over tx. Also, if I should cross multiply, you have 7 times ts equals to 5 times 490.

26:30 The variables side by 7 to the hour can obtain the values I'm looking for, which is the value of

26:35 tx. 7 here can do, 7 here 1, and 7 here 7. Finally, the time at which s diffuses is 5 times 70.

26:48 And that is 350 seconds. I believe you have followed this. Let's see, you can see the time at which

26:55 is 350 seconds. Since you have followed me very well, then you need to be tested also. Assignment.

27:03 The first assignment is a given mass of the gas occupies 150 cm cube at 27 degrees Celsius,

27:08 and a pressure of 1.013 tends to be about 5 Nm square. Calculate the temperature,

27:14 at which its volume will be doubled at the same temperature. You apply the right law,

27:19 you get to answer correctly. The second one says a gas as I used to, a porous portion,

27:25 at the rate of 3 cm cube per seconds. On that same condition, at the rate of 15 seconds,

27:32 15 cents with a cube per seconds. What is the rate of molecular mass of s?

27:36 The third one, which is the last, a given volume of oxygen diffuses through a porous

27:44 partition in 8 seconds. How long will it take? The same volume of carbon, the same volume of carbon

27:50 forzeite to slide views, on that same, given that the atomic mass of carbon is 12 and that of oxygen

27:57 to be seen. Thank you for watching. My name is Adequoli. See you next time.