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Combined and ideal gas laws

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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 :

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

- - 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?

- 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

Avogadro’s Law

- 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

Ideal Gas Law

- 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.

PV

= 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

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 Kelvin
- Volume is in liters
- Pressure is in either atm or mmHg or kPa

L•atm

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

L•atm

.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

L•atm

mol•K

(.0821 )

(1.0atm)

(V)

(8.147L•atm)

=

PV =nRT

(1.0atm)

(V)

=

(273K)

(.3635mol)

V = 8.15L

L•kPa

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

L•kPa

mol•K

(8.31 )

(303kPa)

(V)

(168,970.4 L•kPa)

=

PV = nRT

(303kPa)

(V)

=

(298K)

(68.2 mol)

V = 557.7L

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?

(1.00 atm)

(1.00 L)

nH2O=

(573K)

(.0821L atm/mol K)

nH2O= .021257 mols

2 mol H2

22.4 L H2

=

2 mol H2O

1mol H2

Ideal Gas Law & Stoichiometry

2H2 + O2 2H2O

.021257 mol

.476 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 27C. 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

n = m/MM

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 27C 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 85C and 758 mmHg.

Real Vs. Ideal

- 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.

Real Vs. Ideal

- 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.

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.

- PT=P1+P2+P3+…

- 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 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.

Dalton’s Law of Partial Pressure

- 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.

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 = .285mmHg +

593.525mmHg + PO2

PO2= 167mmHg

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 “contaminated” with water vapor.
- You can determine the pressure of the dry gas by subtracting out the water vapor

Atmospheric

Pressure

Ptot = Patmospheric pressure = Pgas + PH2O

- The water’s vapor pressure can be determined from a list 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.

PH2O at 20.0°C= 17.5 mmHg

Simple Dalton’s Law Calculation

PT = PH2O + PO2

PH2O = 17.5 mmHg

PT = 730 mmHg

730mmHg = 17.5468 + PO2

PO2= 712.5 mmHg

Graham’s Law

- 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

Diffusion

Effusion

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

Graham’s Law

- 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:

Rate of effusion of A

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

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 1.4 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?

40 g

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.

Classroom Practice 2

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

Helium is collected over water @ 25C 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?