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Gas Laws

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**The 4 Gas Law Variables 1. Volume (V) 2. Temperature (T)**

3. Pressure (P) 4. Amount of Gas (n)

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

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**Temperature Kelvin = celcius + 273 (32 oC = 305 K)**

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

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**Conversion Practice 25 + 273 = 298 K Convert 375 K to oC 375 – 273 =**

Convert 25 oC to Kelvin = 298 K Convert 375 K to oC 375 – 273 = 102 oC Convert -50 oC to Kelvin = 223 K

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**Pressure 1. atmosphere (atm) Unfortunately all 4 will be used**

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) 4. torr Unfortunately all 4 will be used

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

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**Pressure Units Conversions**

*Remember* 1 atm = mm Hg = kPa Convert atm to mm Hg 0.875 atm x mm Hg 1 atm Convert 745 mm Hg to atm 745 mm Hg x atm 760 mm Hg Convert atm to kPa 0.955 atm x kPa 1 atm . = 665 mm Hg = atm = 96.8 kPa

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Amount of Gas 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.

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**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) If the volume of a container is increased, the pressure decreases If the volume of a container is decreased, the pressure increases P V or P V

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**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 is decreased, gas molecules have to go a shorter distance to strike the walls, therefore, more collisions, increasing the pressure

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**Boyles Law: The Pressure-Volume Relationship**

The mathematical form of Boyle's Law is: PV = k 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 is an inverse mathematical relationship. As one quantity goes up in the value, the other goes down

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**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. 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.

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**Boyles Law: The Pressure-Volume Relationship**

So we know this: P1V1 = k And we know that the second data pair equals the same constant: P2V2 = k Since k = k, we can conclude that P1V1 = P2V2. This equation of P1V1 = P2V2 will be very helpful in solving Boyle's Law problems.

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**Boyles Law: The Pressure-Volume Relationship**

Example #1: 2.00 L of a gas is at mmHg pressure. What is its volume at standard pressure? ** remember standard pressure = mmHg** Use the equation P1V1 = P2V2 2.00 L x mmHg = mmHg x V2 V2 = 1.95 L

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**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 x 1.08 atm = P2 x L

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**Boyles Law: The Pressure-Volume Relationship**

Example #3: 2.50 L of a gas was at an unknown pressure. However, at standard pressure ( kPa), its volume was measured to be 8.00 L. What was the unknown pressure? insert into P1V1 = P2V2 for the solution.

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**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.

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**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

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**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

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**Charles Law: The Volume-Temperature Relationship**

The mathematical form of Charles' Law is: V / T = k Let V1 and T1 be a volume-temperature pair of data at the start of an experiment. If the volume is changed to a new value called V2, then the temperature must change to T2. 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.

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**Charles Law: The Volume-Temperature Relationship**

So we know this: V1/T1 = k AND And we know this: V2/T2 = k Since k = k, we can conclude that

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**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?

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**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:

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**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?

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**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:

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**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?

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**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:

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**Gay-Lussac’s Law Pressure-Temperature Relationship**

The relationship between pressure and temperature when volume and amount are held constant If the temperature of a container is increased, the pressure increases. If the temperature of a container is decreased, the pressure decreases.

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**Gay-Lussac’s Law Pressure-Temperature Relationship**

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.

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**Gay-Lussac’s Law Pressure-Temperature Relationship**

Gay-Lussac's Law is a direct mathematical relationship. 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

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**Gay-Lussac’s Law Pressure-Temperature Relationship**

Let P1 and T1 be a pressure-temperature pair of data at the start of an experiment. If the temperature is changed to a new value called T2, then the pressure will change to P2. Since k = k, we can conclude that P1 ÷ T1 = P2 ÷ T2

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**Gay-Lussac’s Law Pressure-Temperature Relationship**

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?

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**Gay-Lussac’s Law Pressure-Temperature Relationship**

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 K and remember that standard pressure in kPa is Insert values into the equation

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**Gay-Lussac’s Law Pressure-Temperature Relationship**

Example #2: 5.00 L of a gas is collected at 22.0°C and mmHg. When the temperature is changed to standard, what is the new pressure?

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**Gay-Lussac’s Law Pressure-Temperature Relationship**

Example #2: 5.00 L of a gas is collected at 22.0°C and mmHg. When the temperature is changed to standard, what is the new pressure?

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**Gay-Lussac’s Law Pressure-Temperature Relationship**

Example #2: 5.00 L of a gas is collected at 22.0°C and mmHg. When the temperature is changed to standard, what is the new pressure? = mmHg

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**Avogadro's Law Amount of Gas-Volume Relationship**

the relationship between volume and amount when pressure and temperature are held constant Remember amount is measured in moles

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**Avogadro's Law Amount of Gas-Volume Relationship**

If the amount of gas in a container is increased, the volume increases. n V If the amount of gas in a container is decreased, the volume decreases

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**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

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**Avogadro's Law Amount of Gas-Volume Relationship**

The mathematical form of Avogadro's Law is: V ÷ n = k Let V1 and n1 be a volume-amount pair of data at the start of an experiment. If the amount is changed to a new value called n2, then the volume will change to V2.

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**Avogadro's Law Amount of Gas-Volume Relationship**

We know this: V1 ÷ n1 = k And we know this: V2 ÷ n2 = k Since k = k, we can conclude that V1 ÷ n1 = V2 ÷ n2

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**Avogadro's Law Amount of Gas-Volume Relationship**

Example #1: 5.00 L of a gas is known to contain mol. If the amount of gas is increased to 1.80 mol, what new volume will result

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**Avogadro's Law Amount of Gas-Volume Relationship**

Example #1: 5.00 L of a gas is known to contain mol. If the amount of gas is increased to 1.80 mol, what new volume will result 5.00 L = X 0.965 mol mol

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**Avogadro's Law Amount of Gas-Volume Relationship**

Example #1: 5.00 L of a gas is known to contain mol. If the amount of gas is increased to 1.80 mol, what new volume will result 5.00 L = X 0.965 mol mol = 9.33 L

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The Ideal Gas Law 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.

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**The Ideal Gas Law PV = nRT**

The R in this equation is called the gas constant. R = atm-L/ mol-K atm-L/ mol-K is read as atmosphere liter per mole Kelvin

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**The Ideal Gas Law Example #1**

How many moles of a gas at 100 oC does it take to fill a 1.00 L flask to a pressure of 1.50 atm?

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**The Ideal Gas Law Example #1**

How many moles of a gas at 100 oC does it take to fill a 1.00 L flask to a pressure of 1.50 atm? V = L P = 1.50 atm T = 100 oC convert to K = 373 K n = ? Remember R =

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**The Ideal Gas Law Example #1**

How many moles of a gas at 100 oC does it take to fill a 1.00 L flask to a pressure of 1.50 atm? Solve for n: n = PV/RT (1.50 atm)(1.00 L) ( atm-L/mol-K)(373K)

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**The Ideal Gas Law Example #1**

How many moles of a gas at 100 oC does it take to fill a 1.00 L flask to a pressure of 1.50 atm? Solve for n: n = PV/RT (1.50 atm)(1.00 L) ( atm-L/mol-K)(373K) = mol

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**The Ideal Gas Law Example #2**

What is the volume occupied by 9.45g of C2H2 at STP?

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**The Ideal Gas Law Example #2**

What is the volume occupied by 9.45g of C2H2 at STP? First change grams to moles: 9.45g x 1 mol = moles 26 g

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**The Ideal Gas Law Example #2**

What is the volume occupied by 9.45g of C2H2 at STP? Next solve for V: V = nRT P R = atm-L/mol-K n = moles T = 273 K P = 1 atm

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**The Ideal Gas Law Example #2**

What is the volume occupied by 9.45g of C2H2 at STP? V = (0.363mol)(0.0821atm-L/mol-K)(273 K) 1 atm V = 8.14 L

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Gases Chapter 13.

Gases Chapter 13.

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