The Gas LawsThe Gas LawsThe Gas LawsThe Gas LawsLearning about the special behavior of Learning about the special behavior of

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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.