Gas Laws:

Relationship between Pressure (P), Volume (V), Temperature (T) and quantity; Moles (n)

Boyle’s Law (PV = constant)

This is an inverse relationship: As volume decreases, pressure increases.

Charles’ Law: (V/T = constant)

This is an inverse relationship: As volume decreases, pressure increases.

Charles’ Law: (V/T = constant)

This is a direct relationship. As Temperature decreases, Volume decreases.

Avogadro’s Law: (V/n = constant)

This is a direct relationship. As the number of moles decreases, the volume decreases

Summary:

Combined Gas Law:

PV/nT = constant (T in Kelvins)

P1V1/n1T1 = P2V2/n2T2

Ideal gas Law:

PV = nRT

R = .0821L atm/mol K

Gay-Lussac’s/Avogadro’s Law of Combining Volumes

Equal volumes of any gases at the same temperature and pressure contain the same number of moles of

gas.

The coefficients of a balanced equation can be used to calculate relative volumes.

Standard Molar Volume: At standard temperature and pressure (STP = 1atm and 273.15K) 1 mole of any

ideal gas has a volume of 22.4L

Variations on the ideal gas law equation:

PV = mRT/M (m = sample mass, M = molar mass of the gas)

d = MP/RT (d = density of the gas in g/L)

Examples:

1. Calculate:

a. The new pressure in a closed container if a 5.0L volume of gas at 2.5atm has its volume increased to

7.5L.

b. The new volume of gas (at constant T and P) if 2.0mol of He in a 3.0L container has another 3.0mol of

He placed into the container.

Answers:

a. (5.0L)(2.5atm) = (7.5L)(P2)

P2 = 1.7atm

b. 3.0L/2.0mol = V2/5.0mol

V2 = 7.5L

2. When a rigid hollow sphere containing 680 L of helium gas is heated from 300.K to 600.K, the

pressure of the gas increases to 18atm. How many moles of helium does the sphere contain?

Answer:

n = PV/RT = (18atm)(680L)/(.0821)(600.K)

n = 248.48 = 250moles

3. A child has a lung capacity of 2.2 L. How many grams of air do her lungs hold at a pressure of 1.0

atm and a normal body temperature of 37

o

C? Assume a “formula mass” of 29g/mol for air.

Answer:

m = MPV/RT = (29g/mol)(1.0atm)(2.2L)/(.0821Latm/molK)(310.15K) = 2.5g

4. A gas with a volume of 300.mL at 150.

o

C is heated until its volume is 600.mL. What is the new

temperature of the gas if the pressure is unaltered?

Answer:

300mL/423.15K = 600mL/T2

T2 = 846K = 573

o

C5. Calculate the number of liters occupied, at STP.

a. 0.350 mol O2

b. 63.5g He

Answers:

a. 0.350mol (22.4L/mol) = 7.84L

b. (63.5g)(1mol/4.003g)(22.4L/1mol) = 355L

6. Determine the molar mass of a gas for which a 2.5g sample of that gas occupies a volume of 3.0L at

STP.

Answer:

M =(2.5g)(.0821)(273.15K)/(3.0L)(1atm) =18.7g/mol

or

(2.5g)/(3.0L/22.4L/mol) = 18.7g/mol

7. Find the density of fluorine gas (g/L) at 700torr and 50

o

C.

Answer:

d = MP/RT

= (38.00g/mol)(700/760) / (.0821)(50+273.15) = 1.32g/L

8. For the equation

Ag2S(s)

+ H2(g) → Ag(s)

+ H2S(g)

How many Liters of H2S can be produced from 15.0g of Ag2S and 1.00L of H2(g)

if the reaction occurs at

STP?

Answer:

Ag2S(s)

+ H2(g) → 2Ag(s)

+ H2S(g)

mol Ag2S = 15.0g (1mol/247.8g) = .0605mol

mol H2 = (1.00L)(1mol / 22.4L) = .0446mol

Hydrogen gas limits

0446mol H2 (1mol H2S / 1mol H2) = .0446mol H2S.

V = .0446mol (22.4L/mol) = 1.0L

(Note that the last two steps aren’t really necessary because of the 1:1 mole ratio between H2S and H2.)

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