Entry Task: March Block 2 QUESTION: P= 30 atm V= 50 L T= 293K R= n= X Solve for the number of moles (n)

Agenda Go over Combined and Ideal ws HW: Pre-Lab Proving gas law

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

1. If a gas occupies a volume of 100 cm 3 at a pressure of kPa and 27  C, what volume will the gas occupy at 120 kPa and 50  C? P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 100cm kPa = 300K 120 kPa X cm = 323K

(101.3 kPa)(100cm 3 ) 300 K 323 K = (120 kPa) (X cm 3 ) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 100cm kPa = 300K 120 kPa X cm = 323K

GET X by its self!! (101.3 kPa)(100 cm 3 )(323K) (300 K)(120 kPa) = X cm 3 (101.3 kPa)(100cm 3 ) 300 K 323 K = (120 kPa) (X cm 3 )

DO the MATH cm = 90.9 cm 3 (101.3)(100 cm 3 )(323) (300 )(120) = X cm 3

2. A toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 L. If the temperature where the balloon is released is 20˚ C, what will happen to the volume when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is –15˚ C? P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 5.0 L 1.05 atm = 293K 0.65 atm X L = 258K

(1.05 atm) (5.0L) 293 K 258 K = (0.65 atm) (X L) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 5.0 L 1.05 atm = 293K 0.65 atm X L = 258K

GET X by its self!! (1.05 atm)(5.0L)(258K) (293 K)(0.65 atm) = X L (1.05 atm) (5.0L) 293 K 258 K = (0.65 atm) (X L)

DO the MATH = 7.11L (1.05)(5.0L)(258) (293)(0.65) = X L

3. A closed gas system initially has volume and temperature of 2.7 L and 466 K with the pressure unknown. If the same closed system has values of 1.01 atm, 4.70 L and 605 K, what was the initial pressure in atm? P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 2.7L X atm 466K 1.01 atm 4.70 L 605K

(X atm) (2.7L) 466 K 605 K = (1.01atm) (4.70L) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 2.7L X atm 466K 1.01 atm 4.70 L 605K

GET X by its self!! (466K)(1.01 atm)(4.70L) (2.7L)(605 K) = X atm (X atm) (2.7L) 466 K 605 K = (1.01atm) (4.70L)

DO the MATH = 1.35 atm (466)(1.01 atm)(4.70) (2.7)(605 ) = X atm

4. A closed gas system initially has pressure and temperature of kPa and 692.0°C with the volume unknown. If the same closed system has values of kPa, 7.37 L and °C, what was the initial volume in L? P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = X L kPa = 965K kPa 7.37 L = 225K

(153.3 kPa)(XL) 965 K 225 K = (32.26 kPa) (7.37L) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = X L kPa = 965K kPa 7.37 L = 225K

GET X by its self!! (965K)(32.26kPa)(7.37L) (153.3 kPa)(225 K) = X L (153.3 kPa)(XL) 965 K 225 K = (32.26 kPa) (7.37L)

DO the MATH = 6.65 L (965)(32.26)(7.37L) (153.3)(225) = X L

PV=nRT * Provide the equation for the Ideal gas law.

5. At what temperature (in Kelvin) would 4.0 moles of Hydrogen gas in a 100 liter container exert a pressure of 1.00 atmospheres? P= V= T=R= n= X 100L 1.00 atm mol (1.00 atm)(100L) =(4.0 mol)(0.0821)(X)

Get X by itself! (1.00 atm)(100L) =X (4.0) (0.0821) = 305 K (1.00 atm)(100L) =(4.0 mol)(0.0821)(X)

6. An 18 liter container holds grams of O 2 at 45°C. What is the pressure (atm) of the container? P= V= T=R= n= = 318K 18 L X atm mol (X atm)(18 L) =(0.5 mol)(0.0821)(318K)

Get X by itself! (0.5)(0.0821)(318) =X (18) = 0.73 atm (X atm)(18 L) =(0.5 mol)(0.0821)(318K)

7. How many moles of oxygen must be in a 3.00 liter container in order to exert a pressure of 2.00 atmospheres at 25 °C? P= V= T=R= n= = 298 K 3.00 L 2.00 atm X mol (2.00 atm)(3.00L) =(X mol)(0.0821)(298K)

Get X by itself! (2.00)(3.00) = X (0.0821)(298) = mol (2.00 atm)(3.00L) =(X mol)(0.0821)(298K)

8. A flashbulb of volume L contains O 2 gas at a pressure of 2.3 atm and a temperature of 26  C. How many moles of O 2 does the flashbulb contain? P= V= T=R= n= =299K L 2.3 atm X mol (2.3 atm)(0.0026L) = (X mol)(0.0821)(299)

Get X by itself (2.3)(0.0026) = X (0.0821)(299) = mol of O 2 OR 2.4 x mol (2.3 atm)(0.0026L) = (X mol)(0.0821)(299)

In-class Ch. 14 sec. 3 worksheet

Homework: Combo and ideal #2 ws