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Charles’ Law
WARMUP1.Graph the data2.How are temperature and volume related?3.At what temperature would volume be zero?
volu
me
(cm
3)
Temperature (⁰C)
T and V have a direct, linear relationship
V1 = V2
T1 T2
Temp. (⁰C) Volume (cm3)
30 80-10 70-50 60-90 50-130 40-170 30-210 20-250 10
-273⁰C
V1 = V2
T1 T2
(V1) = 2.5cm3
291.5K 311.5K
Answer: 2.3 cm3
1. Some students think that teachers are full of hot air. If Ms Jonson takes a deep breath of air at 18.5°C and it heats to 38.5°C and the volume expands to 2.5 cm3, what was the starting volume of air? Assume constant pressure.
P1 = cons. P2 = cons.
V1 = ? V2 = 2.5 cm3
T1 = 18.5°C T2 = 38.5°C
291.5 K 311.5 K
2. A sample of nitrogen occupies 250 mL at 25°C. What volume will it occupy at 95°C?
V1 T1
T2
Initial After
P1 = P2 =
V1 = V2 =
T1 = T2 =25°C 298 K
95°C368 K
?250 mL
----- -----
K 298
K 368 X mL 250
V2 = 310 mL
2
2
1
1
TV
TV
1
2 x 12
TTV
V
3. Oxygen gas is at a temperature of 40.0°C when it occupies a volume of 2.30 L. To what temperature (in °C) should it be raised to occupy a volume of 6.50 L?
V1
T1
T2 V2
Initial After
P1 = P2 =
V1 = V2 =
T1 = T2 =40.0°C 313 K
6.50 L
?
2.30 L
----- -----
L 2.30
K313 X L 6.50
T2 = 885 K
2
2
1
1
TV
TV
1
1 x 22
VTV
T
wait…need Celsius -273
T2 = 612°C
Gay-Lussac’s Law
Temperature
Pre
ssur
e
2
2
1
1
TP
TP
How are pressure and temperature related?
P and T have a direct, linear relationship
1. Before a trip from New York to Boston, the pressure of an automobile tire is 1.8 atm at 20.°C. At the end of the trip, the pressure gauge reads 1.9 atm. What is the new Celsius temperature of the air inside the tire? (Assume the tires had a constant volume throughout the whole trip).
Initial After
P1 = P2 =
V1 = V2 =
T1 = T2 =
P1
T1 P2
T2
1.8 atm
1.9 atm
-------- --------
20°C = 293 K ?
2
2
1
1
TP
TP
1
21 2
PPT
T
atm 1.8
atm1.9 X K293
T2 = 309.3 K or 36°C
2. Determine the pressure (in atm) when a constant volume of gas at standard pressure is heated from 20.0°C to 30.0°C. P1T1T2
P2
Initial After
P1 = P2 =
V1 = V2 =
T1 = T2 =20°C
293 K
1.00 atm ?
30°C303 K
----- -----
K293
K303 X atm 1.00
P2 = 1.03 atm
2
2
1
1
TP
TP
1
2 x 12
TTP
P
Rising Water DemoProcedure: Half fill the dish with tap water, add some food coloring, and mix. Light the candle. Place the 250ml beaker over the candle and press it to the bottom of the dish. Observe and explain!
Why does the candle go out?
Lack of oxygen.
Why does the water level rise?
Difference in air pressure causes suction. The cause of the pressure differential is interesting think about the combustion reaction…
The balanced reaction for the combustion of candle wax is given as:
2C20H42(s) + 61O2(g) 40CO2(g) + 42H2O(l)
There is a net loss of 21 moles of gas as O2 is used and CO2 is made. Also, CO2 is more water soluble than O2 (stays dissolved in the liquid rather than being present as gas).
Avogadro’s Law: as moles of gas decreases, volume decreases.
Avogadro’s Law
1 L of Helium
1 L of Oxygen
equal numbers of molecules (moles)
of gas at the same temperature
(with same KE and speed)
V1 = V2
n1 n2
should exert the same pressure,therefore should
have the same volume
1 mole of ANY gas at STP will occupy 22.4 L
molar volume = 22.4 L/mole
1. 4.55 moles hydrogen gas occupy 75.0 mL at STP. Under the same STP conditions, how many moles of gas would be present in a 1680 ml sample?
Initial After
P1 = P2 =
V1 = V2 =
n1 = n2 =
T1 = T2 =
75.0 mL
?
-----
n2 = 102 moles
V1 = V2
n1 n2
----- -----
-----
1680 ml
4.55 moles
n2 = n1V2
V1n2 = (4.55 moles)(1680ml)
(75.0ml)
2. A 5.6 mole sample of air is inside a 2.0L soda bottle. What volume would 0.34 moles of air occupy under the same conditions?
V1n1
V2
Initial After
P1 = P2 =
V1 = V2 =
n1 = n2 =
T1 = T2 =
0.34 moles
?
-----
V2 = 0.12 L
V1 = V2
n1 n2
n2
----- -----
-----
2.0L
5.6 moles
V2 = V1 n2
n1V2 = (2.0L)(0.34 moles)
(5.6 moles)
Gas Law Mystery GameDirections:
You will be given three TIMED gas law problems. (Exactly 3 minutes to do each
problem…so work together). Your job is to find an answer to each mystery problem.
The sum of the answers to all three mystery problems is: -189.088 (ignoring sig fig
rules)
a. You have 60.2 mL of gas. You increase the volume to 701 mL by increasing the temperature to 502°C while holding pressure constant. What was the starting temperature of the gas (in °C)?
b. A 22.0 cm3 sample of oxygen gas expands to 78.0 cm3. If the original pressure was 570. mmHg, what is the new pressure (in atm)?
c. A 1.2 mole sample of acetylene gas has an initial pressure of 699 mm Hg at 40.0°C. When 0.6 moles of acetylene gas are added to the glass container, the volume increases to 25.0 ml. What was the initial volume of acetylene? Assume constant temperature and pressure.