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Gas Laws. The 4 Gas Law Variables 1. Volume (V) 2. Temperature (T) 3. Pressure (P) 4. Amount of Gas (n)

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Presentation on theme: "Gas Laws. The 4 Gas Law Variables 1. Volume (V) 2. Temperature (T) 3. Pressure (P) 4. Amount of Gas (n)"— Presentation transcript:

1 Gas Laws

2 The 4 Gas Law Variables 1. Volume (V) 2. Temperature (T) 3. Pressure (P) 4. Amount of Gas (n)

3 VOLUME Whatever units of volume are used, use them all the way through the problem Symbol = V The 3-dimensional space enclosed by the walls of a container is the volume Usually measured in liters (L) or milliliters (mL)

4 Temperature Kelvin = celcius + 273 (32 o C = 305 K) Standard Temperature is defined as 0 o C or 273 K Symbol = T All gases have a temperature, usually measured in degrees Celsius= o C Also measured in Kelvins = K note there is no degree symbol ALL GAS PROBLEMS WILL BE DONE IN KELVINS

5 Conversion Practice Convert 25 o C to Kelvin 25 + 273 =298 K Convert 375 K to o C Convert -50 o C to Kelvin 375 – 273 = o C 102 o C -50 + 273 =223 K

6 Pressure Unfortunately all 4 will be used Gas Pressure (P) is created by the molecules of a gas colliding with the walls of a container. Gas Pressure (P) is created by the molecules of a gas colliding with the walls of a container. 4 units of pressure 1. atmosphere (atm) 2. millimeters of Mercury (mmHg) 3. kiloPascals (kPa) 3. kiloPascals (kPa) 4. torr

7 Pressure cont…. Standard Pressure is defined as 1 atm or 760 mm Hg or 101.325 kPa or 760 torr Standard Pressure is defined as 1 atm or 760 mm Hg or 101.325 kPa or 760 torr Standard Temperature and Pressure is a common chemistry phrase. It will be abbreviated STP Standard Temperature and Pressure is a common chemistry phrase. It will be abbreviated STP

8 Pressure Units Conversions *Remember* *Remember* 1 atm = 760 mm Hg = 101.325 kPa Convert 0.875 atm to mm Hg 0.875 atm x 760 mm Hg 0.875 atm x 760 mm Hg 1 atm 1 atm Convert 745 mm Hg to atm 745 mm Hg x 1 atm 745 mm Hg x 1 atm 760 mm Hg 760 mm Hg Convert 0.955 atm to kPa 0.955 atm x 101.325 kPa 0.955 atm x 101.325 kPa 1 atm 1 atm. = 665 mm Hg = 0.980 atm = 96.8 kPa

9 Amount of Gas Symbol= n (notice it lower case, all other symbols are upper case) Symbol= n (notice it lower case, all other symbols are upper case) Amount of gas could be measured in moles or grams. If it is in grams you will have to convert to moles at some point. Amount of gas could be measured in moles or grams. If it is in grams you will have to convert to moles at some point.

10 Boyles Law : The Pressure-Volume Relationship Boyles Law shows the inverse relationship between the pressure and the volume of a gas. (Keeping T and n constant) Boyles Law shows the inverse relationship between the pressure and the volume of a gas. (Keeping T and n constant) A. If the volume of a container is increased, the pressure decreases B. If the volume of a container is decreased, the pressure increases P V or P V

11 Boyles Law : The Pressure-Volume Relationship P V or P V WHY ? ? ? ? ? If the volume of the container is increased, gas molecules have to go further to collide with the wall. This means less pressure b/c of less collisions. If the volume of the container is increased, gas molecules have to go further to collide with the wall. This means less pressure b/c of less collisions. If the volume is decreased, gas molecules have to go a shorter distance to strike the walls, therefore, more collisions, increasing the pressure If the volume is decreased, gas molecules have to go a shorter distance to strike the walls, therefore, more collisions, increasing the pressure

12 Boyles Law : The Pressure-Volume Relationship The mathematical form of Boyle's Law is: PV = k The mathematical form of Boyle's Law is: PV = k P = pressure V= volume k= a constant P = pressure V= volume k= a constant This means that the pressure-volume product will always be the same value if the temperature and amount of gas remain constant This means that the pressure-volume product will always be the same value if the temperature and amount of gas remain constant This is an inverse mathematical relationship. As one quantity goes up in the value, the other goes down This is an inverse mathematical relationship. As one quantity goes up in the value, the other goes down

13 Boyles Law : The Pressure-Volume Relationship Suppose P1 and V1 are a pressure-volume pair of data at the start of an experiment. When you multiply P and V together, you get a number that is called k. Suppose P1 and V1 are a pressure-volume pair of data at the start of an experiment. When you multiply P and V together, you get a number that is called k. Now, if the volume is changed to a new value called V2, then the pressure will spontaneously change to P2. It will do so because the PV product must always equal k. Now, if the volume is changed to a new value called V2, then the pressure will spontaneously change to P2. It will do so because the PV product must always equal k.

14 Boyles Law : The Pressure-Volume Relationship So we know this: P 1 V 1 = k So we know this: P 1 V 1 = k And we know that the second data pair equals the same constant: P 2 V 2 = k And we know that the second data pair equals the same constant: P 2 V 2 = k Since k = k, we can conclude that P 1 V 1 = P 2 V 2. Since k = k, we can conclude that P 1 V 1 = P 2 V 2. This equation of P 1 V 1 = P 2 V 2 will be very helpful in solving Boyle's Law problems. This equation of P 1 V 1 = P 2 V 2 will be very helpful in solving Boyle's Law problems.

15 Boyles Law : The Pressure-Volume Relationship Example #1: 2.00 L of a gas is at 740.0 mmHg pressure. What is its volume at standard pressure? ** remember standard pressure = 760.0 mmHg** ** remember standard pressure = 760.0 mmHg** Use the equation P 1 V 1 = P 2 V 2 Use the equation P 1 V 1 = P 2 V 2 2.00 L x 740.0 mmHg = 760.0 mmHg x V 2 V 2 = 1.95 L

16 Boyles Law : The Pressure-Volume Relationship Example #2: 5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is 10.0 L? 5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is 10.0 L? 5.00 L x 1.08 atm = P 2 x 10.0 L

17 Boyles Law : The Pressure-Volume Relationship Example #3: 2.50 L of a gas was at an unknown pressure. However, at standard pressure (101.325 kPa), its volume was measured to be 8.00 L. What was the unknown pressure? insert into P 1 V 1 = P 2 V 2 for the solution. insert into P 1 V 1 = P 2 V 2 for the solution.

18 Charles Law: The Volume-Temperature Relationship This law gives the relationship between volume and temperature if pressure and amount are held constant A. If the temperature increases, the volume of a gas is increased B. If the temperature is decreased, the volume of a gas is decreases.

19 Charles Law: The Volume-Temperature Relationship V T or V T WHY ? ? ? Temperature increase causes molecules to move faster, which makes more collisions with the wall of the container, which increases the volume

20 Charles Law: The Volume-Temperature Relationship Charles' Law is a direct mathematical relationship This means there are two connected values and when one goes up, the other also increases

21 Charles Law: The Volume-Temperature Relationship The mathematical form of Charles' Law is: V / T = k Let V 1 and T 1 be a volume-temperature pair of data at the start of an experiment. If the volume is changed to a new value called V 2, then the temperature must change to T 2. The new volume-temperature data pair will preserve the value of k. The two different volume-temperature data pairs equal the same value and that value is called k.

22 Charles Law: The Volume-Temperature Relationship So we know this: V 1 /T 1 = k AND And we know this: V 2 /T 2 = k Since k = k, we can conclude that

23 Charles Law: The Volume-Temperature Relationship Example #1: A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature?

24 Charles Law: The Volume-Temperature Relationship Example #1: A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature? Answer: Convert 25.0°C to Kelvin and you get 298 K. Standard temperature is 273 K. We plug into our equation like this:

25 Charles Law: The Volume-Temperature Relationship Example #2: 4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C?

26 Charles Law: The Volume-Temperature Relationship Example #2 : 4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C? Answer: Convert 50.0°C to 323 K and 25.0°C to 298 K. Then plug into the equation and solve for x, like this:

27 Charles Law: The Volume-Temperature Relationship Example #3: 5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure?

28 Charles Law: The Volume-Temperature Relationship Example #3: 5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure? Answer:

29 Gay-Lussacs Law Pressure-Temperature Relationship The relationship between pressure and temperature when volume and amount are held constant The relationship between pressure and temperature when volume and amount are held constant A. If the temperature of a container is increased, the pressure increases. B. If the temperature of a container is decreased, the pressure decreases.

30 Gay-Lussacs Law Pressure-Temperature Relationship P T OR P T P T OR P T WHY ? ? ? ? If Temperature is increased, that will cause the molecules to move faster and create more collisions between molecules and the wall of the container. If Temperature is increased, that will cause the molecules to move faster and create more collisions between molecules and the wall of the container.

31 Gay-Lussacs Law Pressure-Temperature Relationship Gay-Lussac's Law is a direct mathematical relationship. Gay-Lussac's Law is a direct mathematical relationship. The mathematical form of Gay-Lussac's Law is: P ÷ T = k The mathematical form of Gay-Lussac's Law is: P ÷ T = k the pressure-temperature fraction will always be the same value if the volume and amount remain constant the pressure-temperature fraction will always be the same value if the volume and amount remain constant

32 Gay-Lussacs Law Pressure-Temperature Relationship Let P 1 and T 1 be a pressure-temperature pair of data at the start of an experiment. If the temperature is changed to a new value called T 2, then the pressure will change to P 2. Let P 1 and T 1 be a pressure-temperature pair of data at the start of an experiment. If the temperature is changed to a new value called T 2, then the pressure will change to P 2. Since k = k, we can conclude that P 1 ÷ T 1 = P 2 ÷ T 2 P 1 ÷ T 1 = P 2 ÷ T 2

33 Gay-Lussacs Law Pressure-Temperature Relationship Example #1: Example #1: 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the pressure to standard pressure?

34 Gay-Lussacs Law Pressure-Temperature Relationship Example #1: Example #1: 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the pressure to standard pressure? change 25.0°C to 298.0 K and remember that standard pressure in kPa is 101.325. Insert values into the equation change 25.0°C to 298.0 K and remember that standard pressure in kPa is 101.325. Insert values into the equation

35 Gay-Lussacs Law Pressure-Temperature Relationship Example #2: Example #2: 5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure?

36 Gay-Lussacs Law Pressure-Temperature Relationship Example #2: Example #2: 5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure?

37 Gay-Lussacs Law Pressure-Temperature Relationship Example #2: Example #2: 5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure? = 689.4 mmHg = 689.4 mmHg

38 Avogadro's Law Amount of Gas-Volume Relationship the relationship between volume and amount when pressure and temperature are held constant the relationship between volume and amount when pressure and temperature are held constant Remember amount is measured in moles Remember amount is measured in moles

39 Avogadro's Law Amount of Gas-Volume Relationship If the amount of gas in a container is increased, the volume increases. If the amount of gas in a container is increased, the volume increases. n V n V If the amount of gas in a container is decreased, the volume decreases If the amount of gas in a container is decreased, the volume decreases n V n V

40 Avogadro's Law Amount of Gas-Volume Relationship WHY ? ? ? If we increase the amount of gas (n) then we have increased the # of molecules. More molecules means more collisions which in-turn means more volume If we increase the amount of gas (n) then we have increased the # of molecules. More molecules means more collisions which in-turn means more volume

41 Avogadro's Law Amount of Gas-Volume Relationship The mathematical form of Avogadro's Law is: V ÷ n = k The mathematical form of Avogadro's Law is: V ÷ n = k Let V 1 and n 1 be a volume-amount pair of data at the start of an experiment. If the amount is changed to a new value called n 2, then the volume will change to V 2. Let V 1 and n 1 be a volume-amount pair of data at the start of an experiment. If the amount is changed to a new value called n 2, then the volume will change to V 2.

42 Avogadro's Law Amount of Gas-Volume Relationship We know this: V 1 ÷ n 1 = k We know this: V 1 ÷ n 1 = k And we know this: V 2 ÷ n 2 = k And we know this: V 2 ÷ n 2 = k Since k = k, we can conclude that V 1 ÷ n 1 = V 2 ÷ n 2 V 1 ÷ n 1 = V 2 ÷ n 2

43 Avogadro's Law Amount of Gas-Volume Relationship Example #1: Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result

44 Avogadro's Law Amount of Gas-Volume Relationship Example #1: Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result 5.00 L = X 5.00 L = X 0.965 mol 1.80 mol 0.965 mol 1.80 mol

45 Avogadro's Law Amount of Gas-Volume Relationship Example #1: Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result 5.00 L = X 5.00 L = X 0.965 mol 1.80 mol 0.965 mol 1.80 mol = 9.33 L = 9.33 L

46 The Ideal Gas Law Describes the behavior of a gas in terms of P, V, n and T Describes the behavior of a gas in terms of P, V, n and T This is a law the encompasses all the other gas laws we have studied. This is a law the encompasses all the other gas laws we have studied.

47 The Ideal Gas Law PV = nRT PV = nRT The R in this equation is called the gas constant. The R in this equation is called the gas constant. R =.0821 atm-L/ mol-K R =.0821 atm-L/ mol-K atm-L/ mol-K is read as atmosphere liter per mole Kelvin atm-L/ mol-K is read as atmosphere liter per mole Kelvin

48 The Ideal Gas Law Example #1 Example #1 How many moles of a gas at 100 o C does it take to fill a 1.00 L flask to a pressure of 1.50 atm?

49 The Ideal Gas Law Example #1 Example #1 How many moles of a gas at 100 o C does it take to fill a 1.00 L flask to a pressure of 1.50 atm? V = 1.00 L P = 1.50 atm T = 100 o C convert to K = 373 K n = ? Remember R = 0.0821

50 The Ideal Gas Law Example #1 Example #1 How many moles of a gas at 100 o C does it take to fill a 1.00 L flask to a pressure of 1.50 atm? Solve for n: n = PV/RT Solve for n: n = PV/RT (1.50 atm)(1.00 L) (1.50 atm)(1.00 L) (0.0821 atm-L/mol-K)(373K) (0.0821 atm-L/mol-K)(373K)

51 The Ideal Gas Law Example #1 Example #1 How many moles of a gas at 100 o C does it take to fill a 1.00 L flask to a pressure of 1.50 atm? Solve for n: n = PV/RT Solve for n: n = PV/RT (1.50 atm)(1.00 L) (1.50 atm)(1.00 L) (0.0821 atm-L/mol-K)(373K) (0.0821 atm-L/mol-K)(373K) = 0.0490 mol = 0.0490 mol

52 The Ideal Gas Law Example #2 Example #2 What is the volume occupied by 9.45g of C 2 H 2 at STP?

53 The Ideal Gas Law Example #2 Example #2 What is the volume occupied by 9.45g of C 2 H 2 at STP? First change grams to moles: 9.45g x 1 mol = 0.363 moles 9.45g x 1 mol = 0.363 moles 26 g 26 g

54 The Ideal Gas Law Example #2 Example #2 What is the volume occupied by 9.45g of C 2 H 2 at STP? Next solve for V: V = nRT P R = 0.0821 atm-L/mol-K n = 0.363 moles T = 273 K P = 1 atm

55 The Ideal Gas Law Example #2 Example #2 What is the volume occupied by 9.45g of C 2 H 2 at STP? V = (0.363 mol )(0.0821 atm-L/mol-K )(273 K ) V = (0.363 mol )(0.0821 atm-L/mol-K )(273 K ) 1 atm 1 atm V = 8.14 L V = 8.14 L


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