Entry Task: March 27-28 Block 2

QUESTION:P= 30 atm V= 50 L T= 293K R= 0.0821 n= X

Solve for the number of moles (n)

Agenda• Go over Combined and Ideal

ws• HW: Pre-Lab Proving gas law

P1 V1 = P2 V2

T1 T2

*Provide the equation for the combined gas law.

1. If a gas occupies a volume of 100 cm3 at a pressure of 101.3 kPa and 27C, what volume will the gas occupy at 120 kPa and 50C?

P1 = V1= T1 =

P2 = V2 = T2 =

100cm3101.3 kPa 27 + 273 = 300K

120 kPa X cm3 50 + 273 = 323K

(101.3 kPa)(100cm3)

300 K 323 K

= (120 kPa) (X cm3)

P1 = V1= T1 =

P2 = V2 = T2 =

100cm3101.3 kPa 27 + 273 = 300K

120 kPa X cm3 50 + 273 = 323K

GET X by its self!!

(101.3 kPa)(100 cm3)(323K)

(300 K)(120 kPa) = X cm3

(101.3 kPa)(100cm3)

300 K 323 K

= (120 kPa) (X cm3)

DO the MATH

3271990 cm3

36000

= 90.9 cm3

(101.3)(100 cm3)(323)

(300 )(120) = X cm3

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?

P1 = V1= T1 =

P2 = V2 = T2 =

5.0 L1.05 atm 20 + 273 = 293K

0.65 atm X L -15 + 273 = 258K

(1.05 atm) (5.0L)

293 K 258 K

= (0.65 atm) (X L)

P1 = V1= T1 =

P2 = V2 = T2 =

5.0 L1.05 atm 20 + 273 = 293K

0.65 atm X L -15 + 273 = 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

1355 190.5 = 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?

P1 = V1= T1 =

P2 = V2 = T2 =

2.7LX atm 466K

1.01 atm 4.70 L 605K

(X atm) (2.7L)

466 K 605 K

= (1.01atm) (4.70L)

P1 = V1= T1 =

P2 = V2 = T2 =

2.7LX 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

22121634

= 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 153.3 kPa and 692.0°C with the volume unknown. If the same closed system has values of 32.26 kPa, 7.37 L and -48.00°C, what was the initial volume in L?

P1 = V1= T1 =

P2 = V2 = T2 =

X L153.3 kPa 692.0+ 273= 965K

32.26 kPa 7.37 L -48.0 + 273= 225K

(153.3 kPa)(XL)

965 K 225 K

= (32.26 kPa) (7.37L)

P1 = V1= T1 =

P2 = V2 = T2 =

X L153.3 kPa 692.0+ 273= 965K

32.26 kPa 7.37 L -48.0 + 273= 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

22943534493 = 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=X100L1.00 atm 0.0821 4.0 mol

(1.00 atm)(100L) =(4.0 mol)(0.0821)(X)

Get X by itself!

(1.00 atm)(100L) =

X(4.0) (0.0821)

1000.3284

= 305 K

(1.00 atm)(100L) =

(4.0 mol)(0.0821)(X)

6. An 18 liter container holds 16.00 grams of O2 at 45°C. What is the pressure (atm) of

the container? 

P= V= T= R= n=

45+273= 318K18 LX atm 0.0821 0.5 mol

(X atm)(18 L) =(0.5 mol)(0.0821)(318K)

Get X by itself!

(0.5)(0.0821)(318) = X(18)

13.118

= 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=

25+273= 298 K

3.00 L2.00 atm

0.0821 X mol

(2.00 atm)(3.00L) =

(X mol)(0.0821)(298K)

Get X by itself!

(2.00)(3.00) = X(0.0821)(298)

6.0024.47

= 0.245 mol

(2.00 atm)(3.00L) =

(X mol)(0.0821)(298K)

8. A flashbulb of volume 0.0026 L contains O2 gas at a pressure of 2.3 atm and a

temperature of 26C. How many moles of O2 does the flashbulb contain?

 P= V= T= R= n=26+273=29

9K0.0026L

2.3 atm 0.0821 X mol

(2.3 atm)(0.0026L) =(X mol)(0.0821)(299)

Get X by itself

(2.3)(0.0026) =X(0.0821)(299)

0.0059824.55

= 0.00024 mol of O2 OR 2.4 x 10-4 mol

(2.3 atm)(0.0026L) =(X mol)(0.0821)(299)

In-class Ch. 14 sec. 3 worksheet

Homework:

Combo and ideal #2 ws