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

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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 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

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 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 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 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?


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