the gas laws learning about the special behavior of gases objective #3 the ideal gas law, pg. 6
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The Gas LawsThe Gas LawsThe Gas LawsThe Gas LawsLearning about the special behavior of Learning about the special behavior of
gasesgases
Objective #3Objective #3The Ideal Gas Law, pg. The Ideal Gas Law, pg.
66
The Ideal Gas Law
• This equation considers a fourth variable, the NUMBER of particles, and the “ideal gas constant” (a.k.a. Avogadro’s Hypothesis) into the combined gas law.
• It would be good to familiarize yourself with the content poster.
The Ideal Gas Law
• This equation considers a fourth variable… the NUMBER of particles!
• This equation incorporates Avogadro’s Hypothesis into the combined gas law.
PV = nRT
Solving for “R”Use the values of standard condition
• A: Using Pressure in kPa
Solving for “R”• A: Using Pressure in kPa
R = Pressure(P) x Volume(V)
# of mol(n)xTemp(K)
Solving for “R”• A: Using Pressure in kPa
R = Pressure(P) x Volume(V) = (101.3kPa)x(22.4L)
# of mol(n)xTemp(K) = (1 mol)x(273K)
Solving for “R”• A: Using Pressure in kPa
R = Pressure(P) x Volume(V) = (101.3kPa)x(22.4L)
# of mol(n)xTemp(K) = (1 mol)x(273K)
R = 8.31
kPa L
Mol K
Solving for “R”• B: Using Pressure in ATM
Solving for “R”• B: Using Pressure in ATM
R = Pressure(P) x Volume(V)
# of mol(n)xTemp(K)
Solving for “R”• B: Using Pressure in ATM
R = Pressure(P) x Volume(V) = (1ATM)x(22.4L)
# of mol(n)xTemp(K) = (1 mol)x(273K)
Solving for “R”• B: Using Pressure in ATM
R = Pressure(P) x Volume(V) = (1ATM)x(22.4L)
# of mol(n)xTemp(K) = (1 mol)x(273K)
R = 0.0821 ATM L
Mol K
Solving for “R”• C: Using Pressure in mm Hg
R = (P) x (V)
(n)x(K)
Solving for “R”•C: Using Pressure in mm Hg
R = (P) x (V) = (760mmHg)x(22.4L) (n)x(K) = (1 mol)x(273K)
Solving for “R”• C: Using Pressure in mm Hg
R = (P) x (V) = (760mmHg)x(22.4L) (n) x (K) = (1 mol)x(273K)
R = 62.4 mmHg L
Mol K
Example 1, pg. 7
• You fill a rigid steel cylinder that has a volume of 20 L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain?
• How many grams is this?
Example 1• You fill a rigid steel cylinder that has a volume of 20
L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain?Rearrange the formula to find what we’re looking for:
• How many grams is this?
Example 1• You fill a rigid steel cylinder that has a volume of 20
L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain?Rearrange the formula to find what we’re looking for:n = PV
RT
• How many grams is this?
Example 1• You fill a rigid steel cylinder that has a volume of 20
L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain?Rearrange the formula to find what we’re looking for:n = PV = (20,000kPa)x(20L) =
RT (8.31 kPaL/molK)x(301K)
• How many grams is this?
Example 1• You fill a rigid steel cylinder that has a volume of 20
L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain?Rearrange the formula to find what we’re looking for:n = PV = (20,000kPa)x(20L) = _________ mol N2
RT (8.31 kPaL/molK)x(301K)
• How many grams is this?
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Example 1• You fill a rigid steel cylinder that has a volume of 20
L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain?Rearrange the formula to find what we’re looking for:n = PV = (20,000kPa)x(20L) = 159.92 mol N2
RT (8.31 kPaL/molK)x(301K)
• How many grams is this?
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Example 1• You fill a rigid steel cylinder that has a volume of 20
L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain?Rearrange the formula to find what we’re looking for:n = PV = (20,000kPa)x(20L) = 159.92 mol N2
RT (8.31 kPaL/molK)x(301K)
• How many grams is this?159.92 mol N2 28 g N2 =
1 1 mol N2
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Example 1• You fill a rigid steel cylinder that has a volume of 20 L
with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain?Rearrange the formula to find what we’re looking for:n = PV = (20,000kPa)x(20L) = 159.92 mol N2
RT (8.31 kPaL/molK)x(301K)
• How many grams is this?159.92 mol N2 28 g N2 = 4,477.7 g
N2
1 1 mol N2
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Example 2• A deep underground cavern contains 2.24 x
106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain?
Example 2• A deep underground cavern contains 2.24 x
106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain?
n = PV RT
Example 2• A deep underground cavern contains 2.24 x
106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain?
n = PV = (1,500kPa)x(2.24 x 106L) RT (8.31 kPaL/molK)x(315K)
Example 2• A deep underground cavern contains 2.24 x
106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain?
n = PV = (1,500kPa)x(2.24 x 106L) = 1.28X106 mol CH4
RT (8.31 kPaL/molK)x(315K)
Example 2• A deep underground cavern contains 2.24 x
106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain?
n = PV = (1,500kPa)x(2.24 x 106L) = 1.28X106 mol CH4
RT (8.31 kPaL/molK)x(315K)
1.28X106 mol CH4 x 16g CH4
1 1 mol CH4
Example 2• A deep underground cavern contains 2.24 x 106 L
of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain?
n = PV = (1,500kPa)x(2.24 x 106L) = 1.28X106 mol CH4
RT (8.31 kPaL/molK)x(315K)
1.28X106 mol CH4 x 16g CH4 = 2.05X107 g CH4
1 1 mol CH4
Example 3. When the temperature of a rigid hollow sphere
containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 10^3 kPa. How many moles of helium does the sphere contain?
Example 3. When the temperature of a rigid hollow sphere
containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 10^3 kPa. How many moles of helium does the sphere contain?
n = PV RT
Example 3. When the temperature of a rigid hollow sphere
containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 10^3 kPa. How many moles of helium does the sphere contain?
n = PV = (1.89x103 kPa)x(685L) RT (8.31 kPaxL/molxK)x(621K)
Example 3. When the temperature of a rigid hollow sphere
containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 10^3 kPa. How many moles of helium does the sphere contain?
n = PV = (1.89x103 kPa)x(685L) = 250.88 moles He
RT (8.31 kPaxL/molxK)x(621K)
Example 4What pressure will be exerted by 0.45 mol of a gas at 25o C if it is contained in a 0.65 L vessel?
Example 4What pressure will be exerted by 0.45 mol of a gas at 25o C if it is contained in a 0.65 L vessel?
P = nRT V
Example 4What pressure will be exerted by 0.45 mol of a gas at 25o C if it is contained in a 0.65 L vessel?
P = nRT = (0.45 mol)x(8.31 kPaL/ molK)x(298K) V 0.65 L
Example 4What pressure will be exerted by 0.45 mol of a gas at 25o C if it is contained in a 0.65 L vessel?
P = nRT = (0.45 mol)x(8.31 kPaL/molK)x(298K) =1714.4 kPa V 0.65 L
Example 5A child has a lung capacity of 2.2 L. How many grams of
air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole.
Example 5A child has a lung capacity of 2.2 L. How many grams of
air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole.
n = PV RT
Example 5A child has a lung capacity of 2.2 L. How many grams of
air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole.
n = PV = __(102kPa)x(2.2L)__ RT (8.31kPaL/molK)x(310K)
Example 5A child has a lung capacity of 2.2 L. How many grams of
air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole.
n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air RT (8.31kPaL/molK)x(310K)
Example 5A child has a lung capacity of 2.2 L. How many grams of
air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole.
n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air RT (8.31kPaL/molK)x(310K)
0.87 mol air x 28g air 1 1 mol air
Example 5A child has a lung capacity of 2.2 L. How many grams of
air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole.
n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air RT (8.31kPaL/molK)x(310K)
0.87 mol air x 28g air = 2.4g air 1 1 mol air
Example 6What volume will 12 grams of oxygen gas occupy at 25o C and a pressure of 52.7 kPa?
Example 6What volume will 12 grams of oxygen gas occupy at 25o C and a pressure of 52.7 kPa?
12g O2 x 1 mol O2 = 0.375 mol O2 1 32g O2
V = nRT = (0.375 molO2)x(8.31kPaL/molK)x(298K) = P (52.7 kPa)
17.62 L O2
It’s always a good idea It’s always a good idea to regularly review the to regularly review the notes we’ve take up to notes we’ve take up to
this point.this point.