In the SI unit system, R = 8.3145 J K-1 mol-1

In the SI unit system, R = 8.3145 J K-1 mol-1 Here, J stands for joule, the SI unit of energy.

In the SI unit system, R = 8.3145 J K-1 mol-1 Here, J stands for joule, the SI unit of energy. 1 J = 1 Nm = 1 kg m2 s-2

In the SI unit system, R = 8.3145 J K-1 mol-1 Here, J stands for joule, the SI unit of energy. 1 J = 1 Nm = 1 kg m2 s-2 Exercise: Try to convert R = 0.082057 l atm K-1 mol-1 to the value given in SI units. (It’s a factor-label exercise).

Problem Example, Ideal Gas Law: Calculate the volume occupied by 0 Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO2 at STP. Assume CO2 can be treated as an ideal gas.

Problem Example, Ideal Gas Law: Calculate the volume occupied by 0 Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO2 at STP. Assume CO2 can be treated as an ideal gas. Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 oC = 273.15 K (exact value)

Problem Example, Ideal Gas Law: Calculate the volume occupied by 0 Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO2 at STP. Assume CO2 can be treated as an ideal gas. Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 oC = 273.15 K (exact value) n R T V = P

Problem Example, Ideal Gas Law: Calculate the volume occupied by 0 Problem Example, Ideal Gas Law: Calculate the volume occupied by 0.168 mol of CO2 at STP. Assume CO2 can be treated as an ideal gas. Given data: n = 0.168 mol, P = 1 atm (exact value) T = 0 oC = 273.15 K (exact value) n R T V = P (0.168 mol) (0.08206 l atm mol-1 K-1) (273 K) (1 atm) = 3.76 l

Combined Gas Law

Combined Gas Law If the initial values of the pressure, volume, temperature, and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write:

Combined Gas Law Pi Vi initial conditions = R ni Ti If the initial values of the pressure, volume, temperature, and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write: Pi Vi initial conditions = R ni Ti

Combined Gas Law Pi Vi initial conditions = R ni Ti Pf Vf If the initial values of the pressure, volume, temperature, and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write: Pi Vi initial conditions = R ni Ti Pf Vf final conditions = R nf Tf

Combined Gas Law Pi Vi initial conditions = R ni Ti Pf Vf If the initial values of the pressure, volume, temperature, and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write: Pi Vi initial conditions = R ni Ti Pf Vf final conditions = R nf Tf Hence: Pi Vi Pf Vf = ni Ti nf Tf

Combined Gas Law Pi Vi initial conditions = R ni Ti Pf Vf If the initial values of the pressure, volume, temperature, and moles are Pi, Vi, Ti, and ni, and if the conditions are changed to a final pressure, volume, temperature, and moles Pf, Vf, Tf, and nf respectively, then we can write: Pi Vi initial conditions = R ni Ti Pf Vf final conditions = R nf Tf Hence: Pi Vi Pf Vf = ni Ti nf Tf This is the Combined Gas Law.

Problem Example, Combined Gas Law: A scuba diver carries three tanks of air. Each has a capacity of 7.00 l and is at a pressure of 1.50 x 102 atm at 25.0 oC. What volume of air does this correspond to at STP?

Problem Example, Combined Gas Law: A scuba diver carries three tanks of air. Each has a capacity of 7.00 l and is at a pressure of 1.50 x 102 atm at 25.0 oC. What volume of air does this correspond to at STP? Given initial data: Pi = 1.50 x 102 atm Vi = 21. 0 l Ti = 298.2 K (25.0 + 273.15)

Problem Example, Combined Gas Law: A scuba diver carries three tanks of air. Each has a capacity of 7.00 l and is at a pressure of 1.50 x 102 atm at 25.0 oC. What volume of air does this correspond to at STP? Given initial data: Pi = 1.50 x 102 atm Vi = 21. 0 l Ti = 298.2 K (25.0 + 273.15) Given final data: Pf = 1 atm Tf = 273.15 K

Pi Vi Pf Vf = ni Ti nf Tf Assume that the number of moles of gas is fixed, so the combined gas law simplifies to: Ti Tf

Pi Vi Pf Vf = ni Ti nf Tf Assume that the number of moles of gas is fixed, so the combined gas law simplifies to: Ti Tf The equation can be rearranged, so that Pi Tf Vf = Vi Pf Ti

(1. 50 x 102 atm) (273. 15 K) Vf = (21. 0 l) (1 atm) (298. 2 K) = 2 (1.50 x 102 atm) (273.15 K) Vf = (21.0 l) (1 atm) (298.2 K) = 2.89 x 103 l at STP

Problem Example: Molar mass of a gas Problem Example: Molar mass of a gas. Calculate the molar mass of methane if 279 ml of the gas measured at 31.3 OC and 492 torr has a mass of 0.116 g. Two steps: first find moles of gas, then determine the molar mass.

Problem Example: Molar mass of a gas Problem Example: Molar mass of a gas. Calculate the molar mass of methane if 279 ml of the gas measured at 31.3 OC and 492 torr has a mass of 0.116 g. Two steps: first find moles of gas, then determine the molar mass. From PV = nRT, PV n = RT

Problem Example: Molar mass of a gas Problem Example: Molar mass of a gas. Calculate the molar mass of methane if 279 ml of the gas measured at 31.3 OC and 492 torr has a mass of 0.116 g. Two steps: first find moles of gas, then determine the molar mass. From PV = nRT, PV n = RT 1 atm 1 l (492 torr)( )( 279 ml)( ) 760 torr 1000 ml = (0.08206 l atm mol-1 K-1)(304.5 K) = 7.23 x 10-3 mol

The molar mass is given by:

The molar mass is given by: = 7.23 x 10-3 mol

The molar mass is given by: = 7.23 x 10-3 mol = 16.0 g mol-1

Dalton’s Law of Partial Pressures

Dalton’s Law of Partial Pressures Dalton’s Law of partial pressures: In a mixture of gases, each component exerts the same pressure as it would, if it were alone and occupied the same volume.

Dalton’s Law of Partial Pressures Dalton’s Law of partial pressures: In a mixture of gases, each component exerts the same pressure as it would, if it were alone and occupied the same volume. Consider a simple case: A mixture of two gases A and B in a container of volume V and temperature T.

The pressure exerted by gas A – called the partial pressure of gas A – is given by:

The pressure exerted by gas A – called the partial pressure of gas A – is given by: nART PA = V

nART PA = V Similarly for gas B, nBRT PB = The pressure exerted by gas A – called the partial pressure of gas A – is given by: nART PA = V Similarly for gas B, nBRT PB =

Now the total pressure PTotal is nTotal RT PTotal = V

Now nTotal = nA + nB so that: Now the total pressure PTotal is nTotal RT PTotal = V Now nTotal = nA + nB so that:

Now nTotal = nA + nB so that: (nA + nB )RT nA RT nB RT PTotal = = + Now the total pressure PTotal is nTotal RT PTotal = V Now nTotal = nA + nB so that: (nA + nB )RT nA RT nB RT PTotal = = + V V V

Now nTotal = nA + nB so that: (nA + nB )RT nA RT nB RT PTotal = = + Now the total pressure PTotal is nTotal RT PTotal = V Now nTotal = nA + nB so that: (nA + nB )RT nA RT nB RT PTotal = = + V V V Hence, PTotal = PA + PB

Now nTotal = nA + nB so that: (nA + nB )RT nA RT nB RT PTotal = = + Now the total pressure PTotal is nTotal RT PTotal = V Now nTotal = nA + nB so that: (nA + nB )RT nA RT nB RT PTotal = = + V V V Hence, PTotal = PA + PB This is Dalton’s Law of partial pressures: The total pressure is the sum of the partial pressures.

The preceding result can be generalized to any number of components: PTotal = PA + PB + PC + PD + … where PA, PB, PC, PD, etc. are the partial pressures of the individual gases.

Problem Example: Assume that 1. 00 moles of air contain 0 Problem Example: Assume that 1.00 moles of air contain 0.78 moles of dinitrogen, 0.21 moles of dioxygen, and 0.01 moles of argon. Calculate the partial pressures of the three gases when the air pressure is at 1.0 atm.

Problem Example: Assume that 1. 00 moles of air contain 0 Problem Example: Assume that 1.00 moles of air contain 0.78 moles of dinitrogen, 0.21 moles of dioxygen, and 0.01 moles of argon. Calculate the partial pressures of the three gases when the air pressure is at 1.0 atm.

Problem Example: Assume that 1. 00 moles of air contain 0 Problem Example: Assume that 1.00 moles of air contain 0.78 moles of dinitrogen, 0.21 moles of dioxygen, and 0.01 moles of argon. Calculate the partial pressures of the three gases when the air pressure is at 1.0 atm.

But

Similar calculations give and

where denotes the vapor pressure of water. Dalton’s law has a practical application when calculating the volume of gases collected over water. For a gas collected over water, the measured pressure is given by where denotes the vapor pressure of water.

First step: Calculate the partial pressure of O2. (Dalton’s law) Problem example: O2 generated in the decomposition of KClO3 is collected over water. The volume of the gas collected at 24 oC and at an atmospheric pressure of 762 torr is 128 ml. Calculate the number of moles of O2 obtained. The vapor pressure of H2O at 24 oC is 22.4 torr. First step: Calculate the partial pressure of O2. (Dalton’s law)

= 762 torr – 22.4 torr = 739.6 torr (extra sig. fig) = 0.973 atm

From the ideal gas equation PV = nRT, (0.973 atm) (0.128 l) = (0.08206 l atm mol-1K-1)(297 K) = 0.00511 mols

Graham’s Law of Effusion

Graham’s Law of Effusion Diffusion: A process by which one gas gradually mixes with another. The term is also used for solutes mixing with a solvent.

Graham’s Law of Effusion Diffusion: A process by which one gas gradually mixes with another. The term is also used for solutes mixing with a solvent. Effusion: The process by which a gas under pressure escapes from one compartment of a container to another by passing through a small opening.

Graham’s law: The rate of effusion of a gas is inversely proportional to the square root of its density when the pressure and temperature are held constant. Effusion rate abbreviated R (don’t get this confused with the gas constant).

Gas density abbreviated d. Graham’s law: The rate of effusion of a gas is inversely proportional to the square root of its density when the pressure and temperature are held constant. Effusion rate abbreviated R (don’t get this confused with the gas constant). Gas density abbreviated d.

Gas density abbreviated d. 1 R (const. P, T) Graham’s law: The rate of effusion of a gas is inversely proportional to the square root of its density when the pressure and temperature are held constant. Effusion rate abbreviated R (don’t get this confused with the gas constant). Gas density abbreviated d. 1 R (const. P, T)

Gas density abbreviated d. 1 R (const. P, T) Graham’s law: The rate of effusion of a gas is inversely proportional to the square root of its density when the pressure and temperature are held constant. Effusion rate abbreviated R (don’t get this confused with the gas constant). Gas density abbreviated d. 1 R (const. P, T) This is Graham’s law of effusion.

The proportionality can be turned into an equality: R = C (const. P and T) This is a statement of Graham’s law of effusion. (The constant is unrelated to any of the previous constants in Boyle’s law, etc.).

For two gases A and B, the relative rates of effusion can be evaluated as follows: