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I. The Gas Laws
I. The Gas Laws
Ch. 14 - GasesCh. 14 - Gases
A. Boyle’s LawA. Boyle’s LawA. Boyle’s LawA. Boyle’s Law
The pressure and volume of a gas are inversely related
• at constant mass & temp
P
V
P1V1 = P2V2
A. Boyle’s LawA. Boyle’s LawA. Boyle’s LawA. Boyle’s Law
V
T
B. Charles’ LawB. Charles’ LawB. Charles’ LawB. Charles’ Law
The volume and absolute temperature (K) of a gas are directly related • at constant mass & pressure
V1 = V2
T1 T2
B. Charles’ LawB. Charles’ LawB. Charles’ LawB. Charles’ Law
V
n
A. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s Principle
Equal volumes of gases contain equal numbers of moles• at constant temp & pressure• true for any gas
V1 = V2
n1 n2
= kPVPTVT
PVnT
D. Combined Gas LawD. Combined Gas LawD. Combined Gas LawD. Combined Gas Law
P1V1
n1T1
=P2V2
n2T2
P1V1T2n2 = P2V2T1n1
GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
E. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law Problems
A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
CHARLES’ LAW
T V
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
E. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law Problems
A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
BOYLE’S LAW
P V
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
GIVEN:
V1 = 2.00 L
n1 = 5.00 moles
n2 = 10.0 moles
V2 = ?
WORK:
V1n2 = V2n1
(2.00 L)(10.0 moles)
=(5.00 moles) V2
V2 = 4.00 L
E. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law Problems A sample of gas occupies 2.00 L with 5.00 moles present.
What would happen to the volume if the number of moles is increased to 10.0?
V n AVOGADRO’S LAW
GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = 101.325 kPa
T2 = 273 K
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=(101.325 kPa) V2 (298 K)
V2 = 5.09 cm3
E. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law Problems
A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.
P T VCOMBINED GAS LAW
Ch. 14 - GasesCh. 14 - Gases
II. Two More LawsII. Two More Laws
B. Dalton’s LawB. Dalton’s LawB. Dalton’s LawB. Dalton’s Law
The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.
Ptotal = P1 + P2 + ...When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
B. Dalton’s LawB. Dalton’s LawB. Dalton’s LawB. Dalton’s Law
Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.
Look up water-vapor pressure on p.899 for 22.5°C.
Sig Figs: Round to least number of decimal places.
The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.
GIVEN:
Pgas = ?
Ptotal = 742.0 torr
PH2O = 42.2 torr
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas?
Look up water-vapor pressure on p.899 for 35.0°C.
Sig Figs: Round to least number of decimal places.
B. Dalton’s LawB. Dalton’s LawB. Dalton’s LawB. Dalton’s Law
The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.
Ideal Gas LawIdeal Gas Law
Ch. 14 GasesCh. 14 Gases
PV
TVn
PVnT
A. Ideal Gas LawA. Ideal Gas LawA. Ideal Gas LawA. Ideal Gas Law
= kUNIVERSAL GAS
CONSTANTR=0.0821 Latm/molK
R=8.315 dm3kPa/molK
= R
You don’t need to memorize these values!
Merge the Combined Gas Law with Avogadro’s Principle:
A. Ideal Gas LawA. Ideal Gas LawA. Ideal Gas LawA. Ideal Gas Law
UNIVERSAL GAS CONSTANT
R=0.0821 Latm/molKR=8.315
dm3kPa/molK
PV=nRT
You don’t need to memorize these values!
GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 LR = 0.0821Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K
P = 3.01 atm
C. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law Problems Calculate the pressure in atmospheres of
0.412 mol of He at 16°C & occupying 3.25 L.
GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = 104.5 kPaR = 8.315 dm3kPa/molK
C. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law ProblemsC. Ideal Gas Law Problems
Find the volume of 85 g of O2 at 25°C and 104.5 kPa.
= 2.7 mol
WORK:
85 g 1 mol = 2.7 mol
32.00 g
PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K
V = 64 dm3
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