e10 the ideal gas law constant, r

January 10 1 Gas Law Exercises

E10 The Ideal Gas Law Constant, R

Dr. Fred O. Garces Chemistry 100 Lab

Miramar College

When Matter Separates


Objective

Use the Ideal Gas Law Equation to compute the Universal Gas Law Constant, “R” and compare it to the theoretical value.


Safety

Handle with Care

After one Trial, HCl will cause this water to be acidic, be careful !


Laws Explaining the Behavior of Gases

Four Law explains the behavior of Ideal Gases*. A. Boyle’s Law P,V (ΔT, Δ n = 0) B. Charles’ Law T,V (Δ P, Δ n = 0) C. Gay-Lussacs’ Law P,T (Δ V, Δ n = 0) D. Avogadro’s Law V, n (Δ P, Δ T = 0) •  Ideal condition: High Temperatures and Low Pressure At Temperature in which condensations occurs, IMF to significant


Summary of Gas Laws

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Boyl " e s Law V = KB

P → P1V1 = P2V2 , ΔT = 0, Δn = 0

Charl " e s Law V = Kc • T → V1

T1 = V2

T2 , ΔP = 0, Δn = 0

Avogadro's Law V = KA • n → V1

n1 = V2

n2 , ΔP = 0, ΔT = 0

V = KB • KC • KA • n TP

KB • KC • KA = R

V = R • n TP

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Ideal Gas Law : PV = nRT

Derivation of R (Universal Gas Constant)

1 mol, T = 0° C, P = 1 atm, V = 22.4 L

R = P• Vn• T

R = 1 atm • 22.4 L1 mol• 273.15 K

R = 0.08206 atm • Lmol• K


Solving Gas Law Problems There are two basic type of gas law problems

Type 1: (Condition-1) Information is provided for a gas with one of

the variable unknown, with the other three known. There is no

change of condition, the problem calls for solving for the unknown

variable. This problem is a straight plug-in to the ideal gas law.

Type2: (Condition-1 / Condition-2) Information is provided for a gas

and a change in one of the four variables causes a change in another

while the two remaining variable is held constant. In this problem

the ideal gas law reduces to one of the individual gas laws. To solve

this problem, the units must be consistent, T must always be in

Kelvin, and R is not involved.


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n1

=1atm• 2.00L

0.08206 L• atmmol• K

• 273K

Using Ideal Gas Law; One condition problem How many moles of xenon are in 2.00L volume, 0 ° C and 1 atm ?

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V1 = 2.00− LT1 = 0°C P1 = 1atm n = ?

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P1V1 = n1 • R • T1

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n1

= 8.93• 10−2mol

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n1

=P1 V1

R • T1


Combine Gas Law: Example: Hydrogen gas (H2) occupies 2.33L at 745 torr and 27°C, what is the new pressure (atm) if the volume is decreased to 1.22L and the temperature is increased to 100°C ?

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P1V1

P2V2 = n1 • R • T1

n2 • R • T2

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P2 =P1V1T2

V2T1

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P1V1

P2V2 = n1 • R • T1

n2 • R • T2

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P1V1

P2V2=

T1

T2

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P1V1T2

V2T1= P2

1

P2=

745 torr • 2.33 L •373 K1.22 L • 300 K

P2=

745 torr • 2.33 L •373 K1.22 L • 300 K

P2= 1769.0torr = 1770 torr

P2= 2.33 atm


Boyle’s Law: Using PV=nRT Example: A 6.50-L sample of oxygen at 24°C is under water at 1.50 atm. If the balloon is moved to a lower depth such that the pressure is 2.50 atm, what is the new volume?

?


Boyle’s Law: Using PV=nRT Example: A 6.50-L sample of oxygen at 24°C is under water at 1.50 atm. If the balloon is moved to a lower depth such that the pressure is 2.50 atm, what is the new volume?

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P1V1

P2V2

= n1 ⋅R ⋅T1

n2 ⋅R ⋅T2

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P1V1

P2V2

= n1 ⋅R ⋅T1

n2 ⋅R ⋅T2

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P1V1 = P2V2

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V2 =P1V1

P2

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V2 =1.50atm⋅6.50L

2.50atm

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V1 = 6.50 −L V1 = ?T1 = 24°C T1 = 24°CP1 =1.50atm P1 = 2.50atm

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V2 = 3.90 L


Charles Law: Using PV=nRT Example: A gas occupies 2.22-L at 0°C. To what temperature must this sample be heated (°C) for it to occupy 3.22 L at constant pressure ?

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V1 = 2.22− L V2 = 3.22− LT1 = 0°C T2 = ?ΔP = 0

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P1V1

P2V2

= n1 ∗ R ∗T1

n2 ∗R ∗T2

P1V1P2V2

= n1 • R • T1n2 • R • T2

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V1

T1

= V2

T2

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T2 = T1V2

V1

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T2 = 273K• 3.22L2.22L

T2 = 396K −273" →"" 123°C


Stoichiometry: Example Example: Hydrogen can be produce by the reaction of 1.33 g of zinc with excess sulfuric acid. What volume (ml) of hydrogen gas is produce at 25°C and 1.12 atm?

Zn (s) + H2SO4(aq) g ZnSO4 (aq) + H2 (g) 1.33 g Excess Vol =?


Stoichiometry Relationship

(g )

( )Volume L

( )Temperature K

Pressure atm( )

R (.0821 atm • L

mol • K)

Density (g / cc)

Molar Mass (g / mol)

MolesABalance equation

← → # # # # # # # # # Stoic. coefficient .

MolesB

Mass (g)

L# of molecules / atoms

Conc. (mol / L )

Vol (L)

Volume (L )

Temperature (K)

Pressure (atm)

N Av (6.02 •1023)

R (. 0821 atm • L

mol • K)

Vol (L)

)Conc. (mol / L

Density (g / cc)

Mass (g)Molar Mass ( g / mol)

Gas phase

Aqueous phase

Solid phase

Liquidphase particle

(atomic) phase

Vol ( )

Gas phase

Aqueous phase

Solid phase

Vol (L)(l)Liquid

phase(atomic) phase

N Av (6.02 •1023) particle

# of molecules / atoms

A

3 4 C


Stoichiometry: Example

Example: Hydrogen can be produce by the reaction of 1.33 g of zinc with excess Hydrochloric acid. What volume (in ml) of hydrogen gas is produce at 25°C and 1.12 atm?

Zn (s) + 2 HCl(aq) g ZnCl2 (aq) + H2 (g) 1.33 g Excess Vol =?

Strategy: Solve for moles of Zn and use this to calculate the moles of H2. Using the ideal gas law, the Vol of H2, can be determined P = 740 torr (0.97 atm), T = 20°C (293 K) Moles Zn: 1.33 g Zn • (1 mole/65.41 g) = 0.0203 mole Zn Moles of H2: 0.0203 moles Zn • (1 mole H2 / 1 mole Zn) = 0.0203 moles H2 Volume H2: V = nRT/P = 0.0203 mole • (0.08206 L•atm/mol K) • 298 K / 1.12 atm

V = .443 L = 443 ml


Dalton’s Law of Partial Pressure (Collecting Gas over Water)

Sum of the partial pressure equals the total pressure

During collision, what is the relative velocity upon impact? Velocity are additive. Likewise forces are additive. Consider three balloons. What is the total pressure if air is combine into one balloon?


Dalton’s Law of Partial Pressure: Experimental conditions

Consider hydrogen gas (H2) being collected over water, at 1 atm. What is the pressure of the H2 gas that is collected ? T = 21°C or 274 K, Patm = 1.0 atm or 760 torr

Pgas = Patm = PH2 + PH2O 760 torr = PH2 + 18.65 PH2 = 760 torr - 18.65orr PH2 = 741.35 torr

Vapor Pressure of Water (torr)

http://intro.chem.okstate.edu/1515SP01/Database/VPWater.html


PreLab Questions


Reading a Barometer

Step 1. Zero the barometer at the bottom. Use the adjusting knob to carefully lower the surface of the merecury reservoir below the tip of zeroing peg. Then slowly raise the level until the tip of the peg just touches the surface of the mercury.

http://www.scivee.tv/node/4129

Step 3. Read the result. Read the fixed scale to one side of the mercury to get the whole number value - in this case 766 mm. On the sliding vernier, find the line that matches one of the lines on the fixed scale. That is the fraction--in this case 4. The final reading here is 766.4 mm, hence an atmospheric pressure of 766.4 mm Hg.

Step 2. Take the reading at the top. Slide the vernier above the top of the

mercury column. Then slowly lower it until the bottom edge of the vernier

is parallel with the top of the mercury meniscus.


Initial Preparation • .


Prepare to React • .

Repeat second trial … Third if data do not agree.


Dump to Waste • .


Data Table

758.24 mmHg

0.078 g

7.96 mL

39.56 mL

19.8 °C

17.319 mmHg


Data Table Pressure of H2 gas:

Moles of H2 gas

Calculation of R: R = PV/nT

Average value of R


Post Lab Questions

e10 the ideal gas law constant, r

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