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.)
