00:00 Good day, everyone. My name is Uchendo E.N. I am your chemistry teacher for today.

00:19 The topic of today's lesson is Gas Law. Objective. At the end of the lesson, the students

00:28 will be able to explain gas law, explain the relationship between absolute temperature

00:34 and the volume of gas at constant pressure, explain the effects of pressure and temperature

00:39 on the volume of gas, perform simple calculation on gas laws accurately.

00:46 Voice Law. Voice Law explains the relationship between volume and fixed mass of a gas. Voice

00:56 Law states that the volume of a given mass of gas is inversely proportional to temperature,

01:04 pressure remaining constant. As we have it on the board, we have V is inversely proportional

01:16 to 1 over P, K is a proportionality constant, that is V is equals to K over P. V is equals

01:27 to K. K, K is a constant of proportionality. Using it to solve calculation, that is using

01:36 two volumes and two pressure. We have P1, V1 is equals to P2, V2. This is a graphical

01:48 representation of voice law. We have P is inversely proportional to V. As we have on the board,

01:59 this is also graphical representation of voice law. Another diagram again is tutorial

02:09 representation of voice law. We also have P is inversely proportional to volume as we

02:17 have on the board. So we have several of P, P1, V1, P2, V2, P3, V3, and so on.

02:31 Chasse Law. The chasse law says that the volume of a given mass of gas is directly proportional

02:41 to its absolute temperature or temperature in Kelvin. You know, we have sessions temperature

02:50 or centigrade temperature. But before you start doing calculations involving temperature,

02:56 all temperatures must be converted to Kelvin temperature. And if you are stating the

03:01 chasse law without putting Kelvin temperature, you will not score. So chasse law says that

03:07 the volume of a given mass of gas is directly proportional to its absolute temperature or

03:13 temperature in Kelvin. Mathematically, V is directly proportional to C. We are pressure

03:22 is constant. And when we are using two volumes and two temperature, we have V1 over T1.

03:29 It's equal to V2 over T2. Temperature. Care is equal to 0 degrees Celsius plus 273.

03:43 So if you are converting Celsius temperature, so centigrade temperature, you have to add

03:49 273 as we have on the board. Chasse law says that the volume of a given mass of gas is

03:59 directly proportional to its absolute temperature or temperature in Kelvin. All temperatures

04:06 must be converted together. You know, we have sessions temperature or centigrade temperature,

04:12 which is 0 degrees Celsius. We use this freezing point of water as a starting point. And the freezing

04:19 point of water is 0 degrees centigrade. But before you solve any calculations involving temperature,

04:25 all temperatures must be converted to Kelvin by adding 273. Mathematically,

04:32 chasse law says that V is directly proportional to C. We are care is constant. And if we are using

04:39 two volumes and two temperature, V1 over T1, it's equal to V2 over T2. Temperature.

04:51 We have Kelvin temperature. So before you solve any calculation involving temperature,

05:10 all temperatures must be converted to Kelvin as we have on the board.

05:15 Kelvin temperature is equal to 0 degrees Celsius plus 273. For example, if you are asked to convert

05:24 12 degrees Celsius to Kelvin temperature is equal to 273 plus 12, which will give you 285 Kelvin.

05:40 Always use absolute temperature. That is Kelvin temperature. When working with gases because

05:46 not every freezing point of 0 degrees centigrade, but motion stops at 0 Kelvin for all substances.

05:55 Using absolute temperature eliminates the use of negative values for temperature.

06:03 Conversion as we have on the board. Conversion, for example, 10 degrees Celsius plus 273,

06:10 125 degrees Celsius plus 273 minus 12 degrees Celsius plus 273. We now have it on the board.

06:18 10 degrees Celsius to Kelvin temperature is 10 plus 273, which will give you 283 Kelvin.

06:27 125 degrees Celsius to Kelvin temperature is 125 plus 273, which will give 298 Kelvin.

06:36 Minus 12 degrees Celsius plus Kelvin temperature is equal to minus 12 plus 273,

06:43 which will give you 2 C1 Kelvin. Graphical representation of Chastula.

06:54 You can have it like this or this way. Where you have minus 273 degrees Celsius, which is equal to

07:01 273 Kelvin.

07:04 Pictorial representation of Chastula. Remember that Chastula V1 over T1 is equal to V2 over T2.

07:15 That's why the pictorial representation as we have on the board, V1 over T1, V2 over T2,

07:24 V3 over T3, V4 over T4 and so on.

07:28 Examples. A gas occupies a volume of 56 CMQ at 1112 millimeter mercury.

07:41 What will be the volume of the gas at 28 millimeter mercury at constant temperature?

07:54 This calculation on the board is on, board is slow. P1, P1, V1 is equal to P2, V2.

08:09 A gas occupies a volume of 56 CMQ at 1112 millimeter mercury. What will be the volume of the gas?

08:19 28 millimeter mercury at constant temperature. So when you have constant temperature,

08:25 you know it is boy slow. Because boy slow states that the volume of a given mass of gas

08:31 is inversely proportional to its pressure, temperature remaining constant.

08:37 So using this one we now have a gas occupies a volume of 56 CMQ. P1 is 112 millimeter mercury.

08:53 V1, 56 CMQ and V1 is what you are looking for. P2 is 28 millimeter mercury. V2 is what you're looking for.

09:18 So making V2 the subject of the formula we have, V2 is equal to P1, V1 over P2,

09:29 substituting in the values we have 112 times by V1, 56,

09:39 all over P2, 28. It's equal to 2, 2, 4, 30 meter Q. It's equal to 2, 2, 2, 4, 30 meter Q.

09:55 So the second example, calculate the temperature of a sample of the gas initially at 75 degree

10:05 stations and the volume of 200 centimeter cube, if the volume increased to 500 CMQ,

10:13 assume the pressure is kept constant. So looking at this question, since the pressure is constant,

10:21 you know it is short slow. Using this relationship, V1 over T1 is equal to V2 over T2.

10:36 V1 equals to 200 centimeter cube, T1 is equal to 75 degree stations for 273 Kelvin. All

10:51 temperatures converted to Kevin and V2 is 500 centimeter cube, T2 is what you're looking for.

11:03 Making T2 the subject of the formula we have, T2 is equal to V2 T1 over V1. Adding 75 to 273,

11:19 the answer is 348 Kelvin. So substituting in the values we have, T2 is equal to

11:30 200 times by 348 all over 500. T2 is equal to V2 T1 over V1, T2 is equal to

11:45 200 times 348 all over 500. It's equal to this result we give us 139.2 Kelvin.

11:57 The relationship between the temperature pressure and volume of a fist mass of gas is obtained

12:06 by combining voice law and chast law. We all know that in voice law, temperature is constant,

12:15 in chast law, pressure is constant. But in general gas equation, all variables are there.

12:23 We have temperature, pressure, volume. So the relationship between temperature,

12:30 pressure and volume of a fist mass of gas is obtained by combining voice law and chast law.

12:37 Voice law, P1 V1 is equal to P2 V2. Chast law, V1 over T1 is equal to V2 over T2.

12:46 Combining the two laws together, we have general gas equation, which all the variables are present.

12:53 P1 over V1, P1 V1 over T1 is equal to P2 V2 over T2.

13:01 Problem 1. A gas occupies 17 cm cube at 71800 Pascal and 20

13:14 and 25 degree sessions. What is the volume of the gas at STP? Using this relationship we have

13:23 P1 V1 over T1 is equal to P2 V2 over T2.

13:31 P1 is 71800 Pascal.

13:43 V1 is the first volume 17 cm cube. T1 is the first temperature,

13:51 25 plus 273. All temperatures converted to Kelvin, which is 298 Kelvin.

14:01 P2 is the standard pressure, which is 101.

14:13 325 Pascal.

14:16 V2 is what you are looking for and T2 is standard temperature, which is 273 Kelvin.

14:30 Making V2 the subject of the formula we have, V2 is equal to P1 V1 T2 over P2 T1.

14:44 V2 is 71800 times V1 17 times T2 to 73 all over P2 101325 times T1 to 9.

15:14 110.36 cm cube.

15:20 Problem 2.

15:23 300 cm cube of a gas was collected at 59 degree sessions and 600 mm of mercury.

15:33 What will be the volume of the gas if the pressure was doubled and the temperature increased to

15:39 97 degrees centigrade. The answer is given as 167 cm cube.

15:46 Ideal gas law. An ideal gas is a gas which obeys boys and chairs law. In practice,

15:56 regas is deviated from those gas laws. The following are the deviations of regas from ideal gas.

16:04 Regas is occupied space. Why are they gas don't? Contrary to the kinetic theory,

16:11 which says that volume of a gas is perfectly negligible. There are forces of attraction between

16:18 molecules of gases. This is because at high pressure and low temperature, gas molecules come together

16:26 and slows down. This makes the volume significant and creates stronger intermolecular forces

16:34 between the molecules. Regas is liquefied at low temperature and high pressure,

16:39 but according to Chasno, ideal gas never liquefied. Conditions under which regas is obeyed the gas laws

16:50 or behave ideally at low pressure and at high temperature. Ideal gas equation. There are four variables

17:02 properties used to measure changes in gases. These are pressure, temperature, volume and number

17:12 of moles. The ideal gas law equation relates all the four variables in the equation in which

17:20 each can vary. This equation is a combination of boys law, Chasno, and Avogadras law. We know that

17:28 our boys law states that the volume of a given mass of gas is inversely proportional to pressure,

17:34 temperature, remaining constants. Why Chasno states that the volume of a given mass of gas is

17:41 directly proportional to its absolute temperature, pressure, remaining constants?

17:57 Ideal gas equation PV is equals to NRC.

18:03 Where R is a gas constant giving us 8.31 views per kilo per mole or 0.82

18:16 atmosphere DMQ per mole per kilo. The value of R to the use depends on the units of

18:24 parameters given. Use 8.31 when pressure is in path, that is Pasca or Newton's per meter speed,

18:34 and volume is in meter cube. You may convert if you like. Use 0.082 when pressure is in atmosphere,

18:44 volume is in DMQ. Well, whenever you are solving any calculation involving ideal gas equation,

18:52 the gas constant to be used will be given to you.

18:55 Ideal gas equation PV is equals to NRT, that you remember that number of moves.

19:13 It's equals to mass over molar mass. So some questions you'll be giving mass and molar mass.

19:28 You will not be giving number of moves at all. So that's what is being explained here.

19:33 That number of moves N equals to mass over molar mass. So between the ideal gas equation,

19:39 we have PV is equals to M, small letter M over capture letter M. That is substituting this number

19:49 of moves. Sometimes in the exam, you will not be giving number of moves. You will be giving mass

19:56 and molar mass. So you know that number of moves is equals to mass over molar mass.

20:04 So you not replace the equation as PV is equals to N for mass given to you.

20:12 Capital letter N for molar mass, NRT. So it is still dissenting as PV is equals to NRT.

20:24 Remember that whenever you are writing, it must be small letter N, not capital letter N for

20:32 number of moves. The problem on ideal gas equation 1, calculate the pressure of two moves of

20:41 a gas which occupies 250 centimeter cube at 10 degrees Celsius. Using the relationship we have

20:49 PV is equals to NRT.

20:57 Where P1, P is equals to P is what you are looking for, V volume 250 centimeter cube.

21:12 N number of moves is equals to two moves, R is a gas constant given to us as

21:28 0.082

21:30 Atmosphere dm cube per kilo per mole.

21:41 C, 10 degrees Celsius, all temperature converted to Kelvin plus 273, 283 Kelvin.

21:57 Making P a subject of the formula we have P is equals to NRT all over V, which is

22:08 N is 2, R 0.082, T 283, all over V 250.

22:23 This will give us 0.186 atmosphere.

22:33 Problem 2, calculate the number of moves present in 200 cm cube of a gas at 25 degrees Celsius

22:41 and 202 kilo pasca. This problem you will do it on your own.

22:47 Evaluation, define chast law, voice law and give the mathematical expression for each of them.

23:01 Give the relationship between volume and pressure as expressed by voice law.

23:07 Thanks for watching. My name is Huchen to EM. This is till next time.