1 / 49

# Combined and ideal gas laws - PowerPoint PPT Presentation

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 :. PV=k 1. V/T=k 2. P/T=k 3. P 1 V 1.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Combined and ideal gas laws' - may-ryan

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Combined and ideal gas laws

• 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 :

PV=k1

V/T=k2

P/T=k3

P1V1

P2V2

=

T1

T2

Combined gas law

• Amount is held constant

• Is used when you have a change in volume, pressure, or temperature

P1V1T2 = P2V2T1

• - 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?

• 1atm

• 2.0 atm

• ?

• 4.0 L

• 273K

• 30°C + 273

• = 303K

P1V1

P2V2

=

T1

T2

Example problem

(1 atm)

(4.0L)

(2 atm)

( V )

2

=

(273K)

(303K)

2.22L = V2

• 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 V & n.

• It turns out that they are directly related to each other.

• As # of moles increases then V increases.

V/n = k

• Which leads us to the ideal gas law –

• The fourth and final variable is amount

• We have been holding it constant.

• We can set up a much more powerful eqn, which can be derived by combining the proportions expressed by the previous laws.

= R

nT

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

PV = nRT

• 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 Kelvin

• Volume is in liters

• Pressure is in either atm or mmHg or kPa

L•kPa

R=.0821

R=8.314

L•mmHg

R=62.4

mol•K

mol•K

mol•K

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

.0821

mol•K

9.45g

26g

Using the Ideal Gas Law

What volume does 9.45g

of C2H2 occupy at STP?

• 1atm

• P

• R

• ?

• V

• T

• 273K

• n

• = .3635 mol

mol•K

(.0821 )

(1.0atm)

(V)

(8.147L•atm)

=

PV =nRT

(1.0atm)

(V)

=

(273K)

(.3635mol)

V = 8.15L

8.31

mol•K

3000g

44g

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?

• 303kPa

• P

• R

• ?

• V

• T

• 298K

• n

• = 68.2 mol

mol•K

(8.31 )

(303kPa)

(V)

(168,970.4 L•kPa)

=

PV = nRT

(303kPa)

(V)

=

(298K)

(68.2 mol)

V = 557.7L

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?

(1.00 atm)

(1.00 L)

nH2O=

(573K)

(.0821L atm/mol K)

nH2O= .021257 mols

22.4 L H2

=

2 mol H2O

1mol H2

Ideal Gas Law & Stoichiometry

2H2 + O2 2H2O

.021257 mol

.476 L H2

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?

• 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

n = m/MM

• We can then use the MM equation to derive a version that solves for the density of a gas.

• Remember that D = m/V

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.

• All of our calculations with gases have been assuming ideal conditions and behaviors.

• We assumed that there was no attraction established between particles.

• We assumed that each particle has no volume of its own

• Under normal atmospheric conditions gases tend to behave as we expect and as predicted by the KMT.

• However, under high pressures and low temperatures, 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.

• 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.

• States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.

• PT=P1+P2+P3+…

• What that means is that each gas involved in a mixture exerts an independent pressure on its container’s walls

• Therefore, to find the pressure in the system you must have the total pressure of all of the gases involved.

• This becomes very important for people who work at high altitudes like mountain climbers and pilots.

• For example, at an altitude of about 10,000m air pressure is about 1/3 of an atmosphere.

• The partial pressure of oxygen at this altitude is less than 50 mmHg.

• By comparison, the partial pressure of oxygen in human alveolar blood needs to be about 100 mmHg.

• Thus, respiration cannot occur normally at this altitude, and an outside source of oxygen is needed in order to survive.

• 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?

PT = PCO2 + PN2 + PO2

760mmHg = .285mmHg +

593.525mmHg + PO2

PO2= 167mmHg

• Partial pressures are also important when a gas is collected through water.

• Any time a gas is collected through water the gas is “contaminated” with water vapor.

• You can determine the pressure of the dry gas by subtracting out the water vapor

Pressure

Ptot = Patmospheric pressure = Pgas + PH2O

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

• 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.

PH2O at 20.0°C= 17.5 mmHg

PT = PH2O + PO2

PH2O = 17.5 mmHg

PT = 730 mmHg

730mmHg = 17.5468 + PO2

PO2= 712.5 mmHg

• Thomas Graham studied the effusion and diffusion of gases.

• Diffusion is the mixing of gases through each other.

• Effusion is the process whereby the molecules of a gas escape from its container through a tiny hole

Effusion

• 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 bigger the molecule the slower it moves the slower it mixes and escapes.

• Kinetic energy can be calculated with the equation ½ mv2

• m is the mass of the object

• v is the velocity.

• If we work with two different at the same temperature their energies would be equal and the equation can be rewritten as:

½ MAvA2 = ½ MBvB2

• “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 ratio of their velocities.

• Rearranging things and taking the square root would give the eqn:

MB

=

Rate of effusion of B

MA

vA

MB

=

vB

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

• 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 1.4 times faster.

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?

4 g

Graham’s Law Example Calc.

Rate of effusion of He

=

Rate of effusion of Ar

Helium is 3.16 times faster than Argon.

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?