GASES AND THE MOLE - LECTURE NOTES

Lecture Video Link:

H.W.: 1a-1c, 2a-2d


A firm knowledge of the behavior of gases is essential to understanding how gases react. In the last section (gases and their properties) you studied several important gas properties, including pressure, volume, temperature and amount of gas (number of moles).

You know how these properties are related. Pressure and volume, for example, are inversely proportional, while Kelvin temperature and volume are directly proportional.

Once you have "mastered" these relationships, you are ready to increase your understanding of gas behavior by extending and applying what you know.

You will extend what you already know by learning the ideal gas law, which summarizes the other gas laws by combining them into a single statement.

DEFINING THE IDEAL GAS LAW:

 Equation

Constant Variables

 CHARLES'S LAW

  PRESSURE & PARTICLES

 BOYLE'S LAW

 TEMPERATURE & PARTICLES

 AVOGADRO'S PRINCIPLE

 TEMPERATURE AND PRESSURE

If the above variables are combined in a single equation, the result is:

V = constant or

Where "R" is a proportionality constant called the UNIVERSAL GAS CONSTANT. By multiplying both sides of the equation by P, the equation becomes:

PV = nRT

and known as the IDEAL GAS LAW. While Charles's law and Boyle's law relate two properties of gases ( V and T, or P and V ), the ideal gas law relates all four properties; the amount of gas, the volume, the temperature, and the pressure.


THE UNIVERSAL GAS CONSTANT:

Before you can use the ideal gas law, you need to know a value for the gas constant, R. This value can be found experimentally under careful controlled conditions.

By convention, many chemists report gas volumes at standard temperature and pressure conditions, or STP, for short. Standard Temperature and Pressure have been arbitrarily defined (not by Mr. Craig) as a temperature of 00C and a pressure of 1 atm. Under these conditions, measurements performed on several gases show that one mole of ANY gas occupies a volume of 22.4 liters.

This information can be used to calculate the value for R:

Although the units for R look confusing, you should not worry about what they mean. R is simply the gas constant; it does NOT stand for a variable such as V (volume) or T (temperature). The units look complex only because of the way R is calculated.

Chemists gradually converting to the use of SI units prefer to use the alternative value for R. Which uses kPa instead of atm - I like the "old fashion" way... But you may choose which you want to use...


EXAMPLE PROBLEM:

Calculate the pressure of 1.65 grams of helium gas at 16.00C and occupying a volume of 3.25 liters.

First, Rearrange the ideal gas law by dividing both sides by V to solve for P:

Next, make a table of variables so you can organize your work and determine what MUST be calculated:

 Variable:

Given:

Need to Calculate:

 P

 unknown

from ideal gas law

 n

 1.65 grams

number of moles

 T

16.00C

value in K

 R

no

 V

3.25 L

no

Find the number of moles of helium: n =

Convert the temperature from Celsius to Kelvin: T = 16.00C + 2730C = 289 K

Substitute all the known values into the ideal gas equation to solve for P:

A better way to write this equation (so that it is easier to cancel the units):


PRACTICE PROBLEM:

1. Calculate the pressure of 159.53 grams of chlorine gas at 53.50C and occupying a volume of 150 cm3.

 

 

CLICK HERE TO VIEW AN EXPLANATION FOR THE ABOVE QUESTION:

2. Calculate the number of moles of nitrogen gas if the volume is 30 liters, at 107 kPa, and a temperature of 298 K.

 

 

Answers to the Examples:


HOMEWORK PROBLEMS:

1a. What is the volume (in liters) of 2.5 moles of oxygen gas measured at 250C and a pressure of 104.5 kpa?

 

 

 

1b. At what temperature will 0.0100 mole of argon gas have a volume of 275 mL at a pressure of 100.0 kPa?

 

 

 

1c. What is the volume (in mL) of 3.6 moles of argon gas measured at 350C and a pressure of 2.2 atm? (answer: 4.1 x 104 mL)

 

 

Answers to the Homework:


APPLYING THE IDEAL GAS LAW:

One advantage of knowing the ideal gas law is that it can be applied to many types of gas problems, including Charles's law and Boyle's law problems.

Suppose that you are working a problem in which the temperature and number of moles of gas do not change; only the pressure and volume are allowed to vary. From this information,

BOYLE'S LAW can be applied:

Here is how to rearrange the ideal gas law. At some initial conditions of pressure and volume, P1 and V1, the ideal gas law says that this is true:

P1V1 = nRT

Similarly, this relationship is true for the final conditions of pressure and volume, P2 and V2:

P2V2 = nRT

Since the number of moles and the temperature do not change, and since the value of R also remains constant, the product nRT represents the same value in both equations. Using the constant k to represent this constant (nRT), these expressions may be rewritten this way:

P1V1 = k = P2V2

CHARLES'S LAW can be applied:

The form of this expression shows that pressure and volume are inversely related. In the same way, if the pressure and number of moles are held constant, the ideal gas law can be rearranged to give this form:

PV1 = nRT1 (initial conditions)

PV2= nRT2 (final conditions)

Therefore:

or

This is the same form predicted by Charles's law, showing a direct relationship between the two variables, volume and temperature.


EXAMPLE: A sample of oxygen gas has a volume of 7.84 cm3 at a pressure of 71.8 kPa and a temperature of 250C. What will be the volume of the gas if the pressure is changed to 101 kPa and the temperature is changed to 00C?

First, identify the variables used in the problem - volume, temperature and pressure. Recoognize that you are going from an initial set of conditions to a final set of conditions. Since no gas is being added to the system and no gas is removed, you can assume that n is a constant as well as R. Using the ideal gas law you can divide both sides by T and write:

Therefore,

This equation (above) is considered the combined gas law - you will see more of this later...

Next, record what you know from the problem:

 P1 = 71.8 kPa

P2 = 101 kPa

V1 = 7.84 cm3

V2 = ?

T1 = 250C

T2 = 00C

Then be sure to always convert degrees Celsius to Kelvins:

T1 = 250C + 273 = 298 K

T2 = 00C + 273 = 273 K

Substitute the values given in the problem into the equation,

and solve for V2.


PRACTICE PROBLEMS:

1. A sample of hydrogen gas at a temperature of 40 K changes its volume from 50 liters to 30 liters. What is the new temperature after the reduction in volume?

 

 

 

CLICK HERE TO VIEW AN EXPLANATION FOR THE ABOVE QUESTION:

2. A sample of radon gas at 300 kPa and 120 0C, and cools down to 80 0C. What is the new pressure if the number of moles and volume remains the same?

 

 

Answers to the Examples:


HOMEWORK PROBLEMS:

2a. A sample of nitrogen gas has a pressure of 2.50 atm at a temperature of 250C. What temperature is required to increase the pressure to 4.00 atm, assuming the volume is fixed and the amount of gas does not change?

 

 

 

 

2b. What happens to the pressure inside a rigid container if the amount of nitrogen gas increases from 0.500 mole to 0.750 mole? The original pressure is 98.0 kPa, and the temperature remains constant.

 

 

 

 

 

2c. The volume of a sample of gas is 200 mL at 275 K and 0.891 atm. What will the new volume of the gas be at 770C and 98.5 kPa?

 

 

 

 

 

2d. What is the volume occupied by 36.0 grams of water vapor at a temperature of 1250C and a pressure of 102 kPa?

 

 

 

Answers to the Homework:


Lecture Video Link:

H.W.: 3a-3d


MOLECULAR MASS DETERMINATION:

As we have seen from above calculations, it is possible to know how many moles "n" of a gas we have in a fixed sample. Suppose we also measure the mass (lets use m for now to represent mass) of the sample. Now we know both the number of moles and the mass. From this information we can now determine the molecular mass of the sample.

mass in grams = number of moles x molar mass in g / mol

m = n x M

Another way to view this problem could be to substitute the units:

We know we will use:

PV = nRT

The "n" = mass of sample (g) / molar mass (g / mol)

so the equation could be written as:

Rearrange the variables:

Which could also look like:

Units:

and of course, re-arrange the units to look "cleaner":


Practice Problems:

1.) If 0.550 grams of a gas occupies 0.200 L at 0.968 atm and 289 K, what is the molecular mass of the gas?

 

 

CLICK HERE TO VIEW AN EXPLANATION FOR THE ABOVE QUESTION:

2.) If 24.23 grams of a gas occupies 8.000 L at 1.500 atm and 396 K, what is the molecular mass of the gas?

 

 

Answers to the Examples:


GAS DENSITIES:

Gases are much less dense than liquids or solids, and we usually present densities in grams per milliliter (centimeters cubed). For gases our volume unit will be in liters.

If the gas is at STP, then the calculation is "simple". For example, if we were calculating the density for one mole of a sample of nitrogen gas at STP we would:

However, the density of nitrogen is different from the conditions at STP (temperature, pressure). In these conditions we need to use the ideal gas law equation. Recall (from above) that n = m / M to obtain:

Rearrange to:

Solve for m:

and then divide by V to get:

Then, substituting this value of m / V into the equation for gas density, d = m / V, we get:


Practice Problems:

1.) Calculate the density of hexane gas, C6H14, in grams per liter at 35 0C and 1.375 atm. (hint: calculate the molar mass first)

 

 

CLICK HERE TO VIEW AN EXPLANATION FOR THE ABOVE QUESTION:

2.) Calculate the density of oxygen gas in grams per liter at 300 K and 0.966 atm.

 

 

 

Answers to the Examples:


HOMEWORK PROBLEMS:

3a. Which of the following gases would have the greatest density at STP? (explain mathematically (for each) how you got the answer)

i. Cl2

ii. SO3

iii. N2O

iv. PF3

3b. Calculate the density of sulfur dioxide gas in grams per liter at 20 0C and 1.333 atm.

 

 

 

3c. If a gas occupies 2.500 L at 1.500 atm and 278 K, what is the molecular mass of the gas if one has twenty grams of the gas?

 

 

 

3d. If 35.23 grams of a gas occupies 23.20 L at 3.500 atm and 560C, what is the molecular mass of the gas?

 

 

Answers to the Homework:


Lecture Video Link:

H.W.: 4a-4c, 5a-5c


THE IDEAL GAS LAW AND STOICHIOMETRY: (yes, it is your favorite "S" word again!)

Not only can the ideal gas law be used to solve problems similar to the last topic (chapter), but it can also be applied to stoichiometry problems involving gases.

One of the variables in the ideal gas law is the amount of gas (or number of moles).

A quick review of the "four" rules of stoichiometry:

1. Write a BALANCED EQUATION
2. CONVERT UNITS to MOLES
3. Determine the MOLE RATION from the balanced equation
4. CONVERT to the DESIRED UNITS

Therefore, use the balanced equation to find the mole product produced or mole reactant needed - then use the Ideal Gas Law to find . . . ?


EXAMPLE: What volume of carbon dioxide forms when 525 mg of calcium carbonate completely reacts with hydrochloric acid? Assume that the carbon dioxide is formed at a pressure of 1 atm and a temperature of 250C. Water vapor and calcium chloride are also formed.

First, Write the balanced equation (step 1):

CaCO3 + 2 HCl ® CO2 + H2O + CaCl2

The equation reminds you that you will need to calculate the number of moles of calcium carbonate that you start with, and that you may determine the number of moles of carbon dioxide that can be produced.

Second, convert the units to moles (step 2):

Third, determine the mole ratio from the balanced equation (step 3):

1 mol CaCO3 is required to produce 1 mol CO2, adding on to the last equation:

Since we are looking for the number of moles, step 4 is not necessary.

Finally, we are ready to apply the Ideal Gas Law. Change the temperature to the Kelvin scale.

250C + 273 = 298 K

Now, rearrange the Ideal Gas Law to solve for volume, and substitute the known values into the equation:

Since, R has the units (liters ~ atm / mol ~ K), we can place the units (mol ~ K) in the denominator to make the canceling of units easier:


PRACTICE PROBLEMS:

1. What temperature will water vapor form when 20 grams of hydrogen completely reacts with oxygen? The water is formed at a pressure of 95 kPa and a volume of 17 liters.

 

 

 

CLICK HERE TO VIEW AN EXPLANATION FOR THE ABOVE QUESTION:

2. When 75 grams of phosphorus pentachloride decomposes when heated it forms what volume of chlorine gas? The other product is phosphorus trichloride. The temperature is 200 0C and the pressure is 150 kPa.

 

 

 

Answers to the Examples:


HOMEWORK PROBLEMS: Use the equation for the ideal gas law to calculate each of the following volumes.

4a. What volume of carbon dioxide forms at 100C and 99 kPa pressure when 0.50 grams of sodium bicarbonate (NaHCO3) reacts completely with hydrochloric acid (HCl)? The other products are water and sodium chloride.

 

 

 

 

4b. Unlike oxygen, nitrogen (the major component of air) is generally unreactive. Magnesium is one metal that can react with nitrogen directly. What volume of nitrogen, measured at a pressure of 102 kPa and a temperature of 270C, will react with 5.0 grams of magnesium? The product is magnesium nitride.

 

 

 

 

4c. What is the pressure of hexane gas, C6H14, that reacts with oxygen gas to form 200 grams of carbon dioxide and water vapor at a temperature of 1200C in a rigid container with a volume of 25 liters?

 

 

 

Answers to the Homework:


GAS STOICHIOMETRY (CONTINUED): STP and MOLAR VOLUME

Earlier you saw how the value of the ideal gas law constant, R, can be determined using the values of the other variables in the Ideal Gas Law.

The set of variables used to calculate R include the volume under a pressure of 1 atm (101.325 kPa) and 00C (273 K). You might recall that the variable for volume is 22.4 liters.

The volume of one mole of a gas is often called a MOLAR VOLUME.

But unlike molar masses, which depend on the substance being measured, the molar volume of every gas is the same, regardless of identity - provided they are under the SAME conditions.

The molar volumes of gases DO change as the conditions of temperature and pressure change - as we discussed already with Charles's law and Boyle's law.


EXAMPLE: A sample of oxygen gas occupies 5.6 liters at STP. How many moles of oxygen are present? What is the mass of the gas sample?

First, ask yourself, "What am I looking for?":

5.6 liters at STP = ? moles = ? grams at STP

Realize that since you are being asked to find the number of moles, you can use the relationship that 1 mole of a gas has a SPECIFIC MOLAR VOLUME. Once you know the number of moles you can use the molar mass of oxygen to calculate the mass of the sample in grams.

Second, The volume of 1 mole of a gas at STP is 22.4 liters which can be written as:

Next, Realize that there is NO equation to balance (since molar volume is the same for all gases in these conditions) so skip Step 1 (balance the equation) and go to Step 2 (convert what is given to moles):

Also, since there is no balanced equation, we can skip Step 3 (Mole Ratio) and go right to Step 4 (Change to the desired units):

Convert the number of moles to grams, using the molar mass of oxygen gas.

As always, you could have put these together to form one equation:


PRACTICE PROBLEMS:

1. How many grams of carbon dioxide gas forms within a 400 liter container at STP?

 

 

CLICK HERE TO VIEW AN EXPLANATION FOR THE ABOVE QUESTION:

2. What is the volume of 19 grams of fluorine gas at STP?

 

 

Answers to the Examples:


HOMEWORK PROBLEMS:

5a. How many moles of ammonia gas, NH3, are required to fill a volume of 50 liters at STP?

 

 

 

 

5b. What is the mass of 1.00 liter of nitrogen at STP?

 

 

 

 

5c. What is the volume of 15.0 grams of H2S at STP?

 

 

 

Answers to the Homework:


Lecture Video Link:

H.W.: 6a-6c, 7a-7d, 8a-8c


DETERMINING VOLUME RATIOS and THE LIMITING REACTANT

Remember when we talked (last November) about chemical equations in a quantitative sense? You learned that the coefficients in a balanced equation give you the information about the number of molecules, (or formula units), and moles that react.

Let's analyze the table below for the reaction of hydrogen gas and nitrogen gas to produce ammonia gas:

3 H2 (g) + N2 (g) ® 2 NH3 (g)

 Molecule Ratio

 3

1

2

Mole Ratio

3

1

2

 Volume Ratio

3

1

2

The mole ratio can be more easily understood if you consider an example where 1 mole of a gas occupies a volume of 22.4 liters.

Therefore, the volume ratios can be written as:

3(22.4 L) H2 (g) + 1(22.4 L) N2 (g) ® 2(22.4 L) NH3 (g)

Giving a ratio of 67.2 liters: 22.4 liters: 44.8 liters, which reduces to 3:1:2.

This means that the coefficients in a balanced chemical equation not only indicate the ratio of reacting molecules and moles yielding a product but also the ratio of reacting volumes.

This is true only if the reaction involves gases. The coefficients do not tell you anything about the volumes if the substances are solids or liquids.

In many situations, an excess of one or more substances is available for a chemical reaction. Some of these excess substances will therefore be left over when the reaction is complete.

The reaction STOPS as soon as one of the reactants is totally consumed.

The substance that is totally consumed in a reaction is called the LIMITING REACTANT or LIMITING REAGENT because it determines, or limits, the amount of product formed.

The other reactant(s) are often called EXCESS REACTANT(S) or EXCESS REAGENT(S).


EXAMPLE: How many liters of carbon dioxide gas are formed when 4.00 liters of methane gas, CH4, react with 0.500 liters of oxygen gas to form carbon dioxide and water vapor? All gases are measured at the same temperature and pressure.

First, write the balanced equation for the chemical reaction:

CH4 + 2 O2 ® CO2 + 2 H2O

Second, ask yourself: "what do I know about this problem ? "

4.00 liters CH4 + 0.500 liters O2 ® ? liters CO2

From the balanced equation, determine the LIMITING REACTANT from the MOLE RATIO of the REACTANTS. The really cool thing about determining the limiting reactant, it does not matter which reactant you choose - to determine the limiting reactant. Notice the two equations below ( KEEP IN MIND that you ONLY need to do ONE of these equations to determine the limiting reactant! )

Okay, let's TALK our way through this...

If you "did" the equation (above) to determine the limiting reactant - say this to yourself: "If I start with 4.00 liters of methane gas ( CH4 ), I would need AT LEAST 8.00 liters of oxygen gas ( O2 ) to COMPLETELY REACT ALL THE methane. Do I have 8.00 liters of oxygen gas (at the beginning?) - NO !!! Therefore, the oxygen gas is the limiting reactant." In this scenario, methane gas COULD NOT be the limiting reactant because I will reactat all of oxygen gas (with the methane gas) before I use all the methane (with the oxygen gas).

OR (notice this is a BIG OR)

IF you "did" the equation (above) to determine the limiting reactant - say this to yourself: "If I start with 0.500 liters of oxygen gas ( O2 ), I would need AT LEAST 0.250 liters of methane gas ( CH4 ) to COMPLETELY REACT ALL THE oxygen. Do I have at least 0.250 liters of methane gas (at the beginning?) - YES !!! I actually have much more !!! So in this scenario, I will USE ALL the oxygen gas and have EXTRA (left-over) methane gas. Therefore, the oxygen gas is the limiting reactant."

Once you know the limiting reactant - you can then calculate how much of the product "WOULD" be formed by the reactants BY USING THE LIMITING REACTANT amount.


PRACTICE PROBLEM:

1. What volume will water vapor form when 40 liters of hydrogen completely reacts with 25 liters of oxygen? Assume the temperature and pressure remain constant.

 

 

 

Answers to the Examples:

CLICK HERE TO VIEW AN EXPLANATION FOR THE ABOVE QUESTION:


HOMEWORK PROBLEMS:

6a. What volume of carbon dioxide gas forms when one liter of oxygen gas reacts with 350 cm3 of methane, CH4? Assume temperature and pressure remain constant. Water is your other product.

 

 

 

 

6b. Propane, C3H8, burns in excess oxygen to form CO2 and H2O. If 22.5 liters of propane burns, what volume of CO2 forms at the same temperature and pressure?

 

 

 

 

6c. Thirty liters of nitrogen gas reacts with 75 liters of hydrogen gas to form what volume of ammonia, NH3, gas? Assume the temperature and pressure remain the same.

 

 

 

Answers to the Homework:


LIMITING REACTANT DEALING WITH ANY STATE OF MATTER:

As you have seen, chemical reactions involve substances in the solid, liquid, or gaseous state. Earlier this year, you dealt with amounts of solids, liquids, and gases in terms of number of particles, moles, and grams.

You now know how to deal with gases in terms of volume, in the following example, you will see how to combine what you learned about stoichiometry with what you have learned about gases in this chapter.

We are beginning to tie everything together . . .


EXAMPLE: When CaCO3 decomposes, how many grams of CaCO3 are needed to produce 9.00 liters of CO2 measured at STP?

First, Write the Balanced Equation:

CaCO3 (s) ® CO2 (g) + CaO (s)

Second, Write down what you are being asked to find.

? grams of CaCO3 = 9.00 liters of CO2

Next, Recognize that the problem is being carried out at STP. This means you can use the standard molar volume. From this you can determine how many moles of CO2 are produced.

The balanced equation tells you the ratio between the number of moles of CaCO3 reacted and CO2 generated. The value you find for the number of moles of CO2 produced will also be the number of moles of CaCO3 needed.


PRACTICE PROBLEM:

When KClO3 decomposes, how many grams of KClO3 are needed to produce 30.0 liters of O2 measured at STP? Also, KCl is produced.

 

Answers to the Examples:

CLICK HERE TO VIEW AN EXPLANATION FOR THE ABOVE QUESTION:


HOMEWORK PROBLEMS:

7a. What mass of CaCO3 must decompose to produce 250 cm3 of CO2 at STP? Calcium oxide is your other product.

 

 

 

 

7b. How many moles of mercury(II)oxide must decompose to produce 2.00 liters of oxygen gas at STP?

 

 

 

 

7c. How many liters of phosphorus pentachloride gas, PCl5, are formed when 7.0 liters of phosphorus trichloride gas, PCl3, react with 9.0 liters of chlorine gas? All gases are equal temperature and pressure.

 

 

 

 

7d. How many grams of gaseous water can be produced from the combustion of 8.0 liters of hydrogen gas with 5.0 liters of oxygen gas? Assume that the water produced is measured at the same temperature and pressure as the reactants.

 

 

 

Answers to the Homework:


DIFFUSION and EFFUSION:

In a sample of air, the molecules of a gas have a wide distribution of speeds. We discussed earlier that average molecular speed translates to the average translational kinetic enery (temperature). The average speed is defined as:

For most applications the root-mean-square speed is a more useful expression of molecular speed than the average speed. The root-mean-square speed, u rms, is the square root of the average of the squares of the speeds of a large number (N) of molecules.

Since we cannot measure the speeds of individual molecules, we cannot use the equations above and thanks to Newton (who invented Calculus) we (well not us) can derive the above equations to the simpler equation below:

Typical u rms speeds of molecules are quite high. For hydrogen at 25 0C, u rms = 1.92 x 103 m/sec (about 4300 mph).

To determine u rms in meters per second with this equation requires that we express the gas constant, R, as 8.314 joules (J) per mol per kelvin ( J mol-1 K-1 ) and the molar mass, M, in kilograms per mole.

Recall, one Joule equals:

So when we plug all the units into the equation:

Are you kidding ???

When all the units cancel, the only units that remain are meters squared per second squared, and when you take the square root of those units you have a speed unit of meters per second - simple... Ha!!!

Looking at the graphic above, as the temperature of a substance increases the distribution of kinetic energy (temperatures) for the substance is "spread out". This will be an issue later this semester when we discuss reaction rates and activation energy.


PRACTICE PROBLEM:

What is the "root-mean-square speed" for propane gas, C3H8, at room temperature, 20 0C?

 

 

Answers to the Examples:

CLICK HERE TO VIEW AN EXPLANATION FOR THE ABOVE QUESTION:


DIFFUSION is defined as the process in which one substance mixes with one or more other substances as a result of translational motion of molecules. Gaseous diffusion is realatively rapid because the molecules are far apart. Diffusion of liquids is much slower than gases, and solids very slow.

EFFUSION is defined as the process in which a gas escapes from its container through a tiny hole, or orifice. Lighter molecules have greater average speeds and we would expect them to escape more quickly through a hole than heavier molecules.

We can relate effusion rates of gases to their root-mean-square speeds. For two gases, 1 and 2, whose molar masses are M1 and M2, the effusion rates are related by the expression:


PRACTICE PROBLEM:

1.) If compared under the same conditions, how much faster than chlorine gas does hydrogen effuse through a tiny hole?

 

 

 

 

Answers to the Examples:

CLICK HERE TO VIEW AN EXPLANATION FOR THE ABOVE QUESTION:


HOMEWORK PROBLEMS:

8a. What is the "root-mean-square speed" for butane gas, C4H10, at 350 K?

 

 

 

8b. If compared under the same conditions, how much faster than radon gas, Rn, does fluorine gas, F2, effuse through a tiny hole?

 

 

8c. Calculate the "root-mean-square-speed" for chlorine gas at room temperature, 20 0C.

 

Answers to the Homework


link to the Extra Practice ; link to the video explation