Entry Task: March 27-28 Block 2 QUESTION: P= 30 atm V= 50 L T= 293K R= 0.0821 n= X Solve for the...
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Transcript of Entry Task: March 27-28 Block 2 QUESTION: P= 30 atm V= 50 L T= 293K R= 0.0821 n= X Solve for the...
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