Combined and ideal gas laws

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Combined and ideal gas laws. Combined gas law. If we combine all of the relationships from the 3 laws covered thus far (Boyle’s, Charles’s, and Gay-Lussac’s) we can develop a mathematical equation that can solve for a situation where 3 variables change : . Combined gas law.

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### Combined and ideal gas laws

Combined gas law

• If we combine all of the relationships from the 3 laws covered thus far (Boyle’s, Charles’s, and Gay-Lussac’s) we can develop a mathematical equation that can solve for a situation where 3 variables change :

Combined gas law

• Amount is held constant
• Is used when you have a change in volume, pressure, or temperature

- P1

• - P2 
• - V1 
• - V2 
• - T1 
• - T2 

Example problem

A gas with a volume of 4.0L at STP. What is its volume at 2.0atm and at 30°C?

P1V1

P2V2

=

T1

T2

Example problem

=

• So far we’ve compared all the variables except the amount of a gas (n).
• There is a lesser known law called Avogadro’s Law which relates ____.
• It turns out that they are _________ related to each other.
• As ____________ increases then V increases.

V/n = k

Ideal Gas Law

• Which leads us to the ideal gas law –
• The fourth and final variable is amount
• ___________________________.
• We can set up a much more powerful eqn, which can be derived by combining the proportions expressed by the previous laws.

Ideal Gas Law

• If we combine all of the laws together including Avogadro’s Law mentioned earlier we get:

Where R is the universal gas constant

Normally

written as

Ideal Gas Constant (R)

• R is a constant that connects the 4 variables
• R is dependent on the units of the variables for P, V, & T
• Temp is always in ________
• Volume is in ________
• Pressure is in either ____________ ____________

Ideal Gas Constant

• Because of the different pressure units there are 3 possibilities for our ideal gas constant
• If pressure is given in atm
• If pressure is given in mmHg
• If pressure is given in kPa

Using the Ideal Gas Law

What volume does 9.45g

of C2H2 occupy at STP?

• P
• R
• V
• T
• n

L•atm

mol•K

(____ )

(___atm)

(V)

(_________)

=

PV =nRT

(______)

(V)

=

(___K)

(_______)

V = ______L

A camping stove propane tank holds 3000g of C3H8. How large a container would be needed to hold the same amount of propane as a gas at 25°C and a pressure of 303 kpa?

• P
• R
• V
• T
• n

L•kPa

mol•K

(____ )

(___kPa)

(V)

(_________ L•kPa)

=

PV = nRT

(______)

(V)

=

(___K)

(____ mol)

Classroom Practice

Use the Ideal Gas Law to complete the following table for ammonia gas (NH3).

PV = nRT

Ideal Gas Law & Stoichiometry

What volume of hydrogen gas must be burned to form 1.00 L of water vapor at 1.00 atm pressure and 300°C?

(____ atm)

(___ L)

nH2O=

(___K)

(____L atm/mol K)

nH2O= _______ mols

_ mol H2

____ L H2

=

_ mol H2O

1mol H2

Ideal Gas Law & Stoichiometry

2H2 + O2 2H2O

.021257 mol

_____ L H2

Classroom Practice

To find the formula of a transition metal carbonyl, one of a family of compounds having the general formula Mx(CO)y, you can heat the solid compound in a vacuum to produce solid metal and CO gas. You heat 0.112 g of Crx(CO)y

Crx(CO)y(s)  x Cr(s) + y CO(g)

and find that the CO evolved has a pressure of 369 mmHg in a 155 ml flask at 27C. What is the empirical formula of Crx(CO)y?

Variations of the Ideal Gas Law

• We can use the ideal gas law to derive a version to solve for MM.
• We need to know that the unit mole is equal to m ÷ MM, where m is the mass of the gas sample

PV = nRT

Variations of the Ideal Gas Law

• We can then use the MM equation to derive a version that solves for the density of a gas.
• Remember that D = m/V

Classroom Practice 1

A gas consisting of only carbon and hydrogen has an empirical formula of CH2. The gas has a density of 1.65 g/L at 27C and 734 mmHg. Determine the molar mass and the molecular formula of the gas.

Silicon tetrachloride (SiCl4) and trichlorosilane (SiHCl3) are both starting materials for the production of electro-nics-grade silicon. Calculate the densities of pure SiCl4 and pure SiHCl3 vapor at 85C and 758 mmHg.

Real Vs. Ideal

• All of our calculations with gases have been assuming ideal conditions and behaviors.
• We assumed that there was ____ __________ established between particles.
• We assumed that each particle has _____________ of its own.
• Under normal atmospheric conditions gases tend to behave as we expect and as predicted by the KMT.

Real Vs. Ideal

• However, under ______________ and _______________, gases tend to deviate from ideal behaviors.
• Under extreme conditions we tend to see a tendency of gases to not behave as independently as the ideal gas law predicts.
• Attractive forces between gas particles under high pressures or low temperature cause the gas not to behave predictably.

Loose Ends of Gases

• There are a couple more laws that we need to address dealing with gases.
• Dalton’s Law of Partial Pressures
• Graham’s Law of Diffusion and Effusion.

Dalton’s Law of Partial Pressure

• States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
• What that means is that each gas involved in a mixture exerts an independent pressure on its container’s walls

Dalton’s Law of Partial Pressure

• Therefore, to find the pressure in the system you must have the ____ ______________of all of the gases involved.
• This becomes very important for people who work at _____ altitudes like mountain climbers and pilots.
• For example, at an altitude of about 10,000m air pressure is about ____ of an atmosphere.

Dalton’s Law of Partial Pressure

• The partial pressure of oxygen at this altitude is less than ___ mmHg.
• By comparison, the partial pressure of oxygen in human alveolar blood needs to be about _____ mmHg.
• Thus, respiration cannot occur normally at this altitude, and an outside source of oxygen is needed in order to survive.

Simple Dalton’s Law Calculation

• Three of the primary components of air are CO2, N2, and O2. In a sample containing a mixture of these gases at exactly 760 mmHg, the partial pressures of CO2 and N2 are given as PCO2= 0.285mmHg and PN2 = 593.525mmHg. What is the partial pressure of O2?

Simple Dalton’s Law Calculation

PT = PCO2 + PN2 + PO2

760mmHg = ______ mmHg +

_________ mmHg + P__

PO2= _____mmHg

Dalton’s Law of Partial Pressure

• Partial pressures are also important when a gas is collected through water.
• Any time a gas is collected through water the gas is “____________” with water vapor.
• You can determine the pressure of the dry gas by __________ out the water vapor

Atmospheric

Pressure

Ptot = Patmospheric pressure = Pgas + PH2O

• The water’s vapor pressure can be determined from ___ and subtract-ed from the atmospheric pressure

Simple Dalton’s Law Calculation

• Determine the partial pressure of oxygen collected by water displace-ment if the water temperature is 20.0°C and the total pressure of the gases in the collection bottle is 730 mmHg.

Simple Dalton’s Law Calculation

PT = PH2O + PO2

PH2O = ____ mmHg

PT = ___ mmHg

___mmHg = _______ + PO2

PO2= _____ mmHg

Graham’s Law

• Thomas Graham studied the _______ and ________ of gases.
• __________ is the mixing of gases through each other.
• _______ is the process whereby the molecules of a gas escape from its container through a tiny hole

Graham’s Law

• Graham’s Law states that the rates of effusion and diffusion of gases at the same temperature and pressure is dependent on the size of the molecule.
• The _________ the molecule the slower it moves the __________ it mixes and escapes.

Graham’s Law

• Kinetic energy can be calculated with the equation ___________
• __ is the mass of the object
• __ is the velocity.
• If we work with two different gases at the same ______________ their energies would be equal and the equation can be rewritten as:

“M” represents molar mass

• “v” represents molecular velocity
• “A” is one gas
• “B” is another gas
• If we want to compare both gases velocities, to determine which gas moves faster, we could write a ____ of their velocities.
• Rearranging things and taking the ____________ would give the eqn:

Rate of effusion of A

MB

=

Rate of effusion of B

MA

• This shows that the velocities of two different gases are inversely propor-tional to the square roots of their molar masses.
• This can be expanded to deal with rates of diffusion or effusion

Graham’s Law

• The way you can interpret the equation is that the number of times faster A moves than B, is the square root of the ratio of the molar mass of B divided by the Molar mass of A
• So if A is half the size of B than it effuses or diffuses ____ times faster.

Rate of effusion of A

MB

=

Rate of effusion of B

MA

Graham’s Law Example Calc.

If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster?

__ g

_ g

Graham’s Law Example Calc.

Rate of effusion of He

=

Rate of effusion of Ar

Helium is ____ times faster than Argon.

Classroom Practice 2

A mixture of 1.00 g H2 and 1.00 g He is placed in a 1.00 L container at 27C. Calculate the partial pressure of each gas and the total pressure.

Helium is collected over water @ 25C and 1.00 atm total pressure. What total volume of gas must be collected to obtain 0.586 g of He?

The rate of effusion of a gas was meas-ured to be 24.0 ml/min. Under the same conditions, the rate of effusion of pure CH4 gas is 47.8 ml/min. What is the molar mass of the unknown gas?