* Discuss Ch. 14 sec. 1-2 ws * Ch. 14 sec. 3 – Combo and Ideal gas law * HW: Combined and Ideal ws.

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* Discuss Ch. 14 sec. 1-2 ws * Ch. 14 sec. 3 – Combo and Ideal gas law * HW: Combined and Ideal ws

* Use the combined gas law equation with gas problems * P 1 V 1 /T 1 = P 2 V 2 /T 2 * Relate the amount of gas to its pressure, temperature and volume by using the ideal gas law * PV = nRT

What does the combined gas law state? Combined gas law states the relationship among pressure, volume, and temperature of a fixed amount of gas.

P 1 V 1 = P 2 V 2 T1T1 T2T2 * Provide the equation for the combined gas law.

* 1. A helium- filled balloon at sea level has a volume of 2.1 L at atm and 36°C. If it is released and rises to an elevation at which the pressure is atm and the temperature is 28°C, what will be the new volume of the balloon. **Remember to change Celsius to kelvin. P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 2.1 L atm = 309K atm X L = 301K

(0.998 atm) (2.1L) 309 K 301 K = (0.900 atm) (X L) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 2.1 L atm = 309K atm X L = 301K

(0.998 atm)(2.1L)(301K) (309 K)(0.900 atm) = X L (0.998 atm) (2.1L) 309 K 301 K = (0.900 atm) (X L)

L = 2.3 L (0.998)(2.1L)(301) (309 )(0.900) = X L

* 2. At 0.00°C and 1.00 atm pressure, a sample of gas occupies 30.0 mL. If the temperature is increased to 30.0°C and the entire gas sample of transferred to a 20.0 mL container, what will be the gas pressure inside the container? P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 30.0 mL 1.00 atm = 273K X atm 20.0 mL = 303K

(1.00 atm) (30.0L) 273 K 303 K = (X atm) (20.0 L) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 30.0 mL 1.00 atm = 273K X atm 20.0 mL = 303K

(1.00 atm)(30.0L)(303K) (273 K)(20.0 L ) = X atm (1.00 atm) (30.0L) 273 K 303 K = (X atm) (20.0 L)

9090 atm 5460 = 1.66 atm (1.00 atm)(30.0)(303) (273)(20.0) = X atm

* 3. A sample of air in a syringe exerts a pressure of 1.02 atm at a temperature of 22.0°C. The syringe is placed in a boiling water bath at 100°C. The pressure of the air is increased to 1.23 atm by pushing the plunger in, which changes the volume to mL. What was the original volume of the air? P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = X mL 1.02 atm = 295K 1.23 atm mL = 373K

(1.02 atm) (XmL) 295 K 373 K = (1.23atm) (0.224 mL) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = X mL 1.02 atm = 295K 1.23 atm mL = 373K

(295K)(1.23 atm)(0.224 mL) (1.02 atm)(373 K) = X mL (1.02 atm) (XmL) 295 K 373 K = (1.23atm) (0.224 mL)

81.3 mL = mL (295)(1.23)(0.224 mL) (1.02)(373) = X mL

State the Avogadro’s principle. It states that equal volumes of gases at the same temperature and pressure contains equal number of particles

Define molar volume. One mole of gas contains 6.02 x10 23 molecules at STP

What is STP? Standard temperature 0.00°C and pressure 1.00 atm.

What is the conversion factor between moles and volumes of gas? 22.4 L /1 mole

1. Determine the volume of a container that holds 2.4 mol of gas at STP? 2.4 mol 1 mol 22.4 L = 54 L

2. What size container do you need to hold mol of N 2 at STP? mol 1 mol 22.4 L = L

What is the fourth variable? The number of moles is the fourth variable

How will changing the number of gas molecules in a system change the other variables? By changing the number of moles of gas, changes the other variables.

Provide the equation for the ideal gas law. Explain what each term means. PV=nRT P is pressure V is volume n is the number of moles R is a gas constant T is temperature in Kelvin

What is the “R” value for: atm___________________________ kPa____________________________ mmHg_________________________ L atm/mol-K L kPa/mol-K 62.4 L mmHg/mol-K

What does the term “ideal gas” mean? An ideal gas is one whose particles take up no space and have no intermolecular forces An ideal gas follows all the gas laws under all pressures and temperatures

When is the ideal gas law NOT likely to work for a real gas? At high temperatures and pressures, real gases do not follow the gas laws.

4. If the pressure exerted by a gas at 25°C in a volume of L is 3.81 atm. How many moles of gas are present? P= V= T=R= n= = L 3.81 atm X (3.81 atm)(0.044L) =(X)(0.0821)(298)

(3.81 atm)(0.044L) =(X)(0.0821)(298) (3.81 atm)(0.044L) =X (0.0821)(298) = mols of gas

5. Determine the Celsius temperature of 2.49 moles of gas contained in 1.00-L vessel at a pressure of 143 kPa P= V= T=R= n= X 1.00 L 143 kPa mol (143 kPa)(1.00L) =(2.49 mol)(8.314)(X)

(143)(1.00) =X (2.49)(8.314) = 6.91K subtract 273 = -266°C (143 kPa)(1.00L) =(2.49 mol)(8.314)(X)

6. Calculate the volume that a mol sample of a gas will occupy at 265K and a pressure of atm. P= V= T=R= n= 265 K X L atm mol (0.900 atm)(XL) =(0.323 mol)(0.0821)(265)

(0.323)(0.0821)(265) = X (0.900) = 7.81 L (0.900 atm)(XL) =(0.323 mol)(0.0821)(265)

7. What is the pressure in atmospheres of a mol sample of helium gas at a temperature of 20°C if its volume is L? P= V= T=R= n= =293K L X atm mol (X atm)(0.505L) =(0.108 mol)(0.0821)(293)

(0.108)(0.0821)(293) = X (0.505) = 5.14 atm (X atm)(0.505L) =(0.108 mol)(0.0821)(293)

8. Determine the Kelvin temperature required for mol of gas to fill a balloon to 1.20 L under atm pressure. P= V= T=R= n= X K 1.20 L atm mol (0.988 atm)(1.20L) =( mol)(0.0821)(X)

(0.988)(1.20) = X (0.0470)(0.0821) = 307K (0.988 atm)(1.20L) =( mol)(0.0821)(X)